At What Horizontal Distance From The Target Should The Package Leave The Airplane?

Therefore, the horizontal distance before dropping the package should be L= x (T) = v_0 T = v_0 sqrt {frac {2 h} {g}} approx 2197.8 m L = x(T) = v0 T = v0

How far in front of the island must the airplane release the package?

The airplane must release the package at least 905.06 m in front of the island. What is the package’s horizontal speed when it reaches the level of the island? What is the package’s vertical speed when it reaches the level of the island?

How high above a glacier does a plane need to fly?

A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 190m above the glacier at a speed of 200m/s. How far short of the target should it drop the package?

What is the speed of an aeroplane moving horizontally?

An aeroplane moving horizontally at a speed of 200 m/s and at a height of 8. 0 × 1 0 3 m is to drop a bomb on a target. At what horizontal distance from the target should the bomb be released

How high can a bomb be dropped from an aeroplane?

A bomb is dropped from an aeroplane when it is directly above a target at a height of 1000 m.The aeroplane is moving horizontally with a speed of 500kmh −1. By the how much distance will the bomb miss the target?

When the pilot drops a package and it reaches the ground what is the plane’s horizontal position ignoring air resistance?

The airplane will be directly above the package when it hits the ground.

What is the horizontal speed of the plane?

Since the plane also travels at a constant horizontal velocity of +115 m/s, it remains directly above the falling package.

What forces are acting on the package after it has been dropped from the plane and is in free fall through the air?

As the package falls, it undergoes a vertical acceleration; that is, there is a change in its vertical velocity. This vertical acceleration is attributed to the downward force of gravity which acts upon the package.

How far does the plane have to fly to land?

But the typical “cruising altitude” – that is, the highest altitude reached during a flight and sustained between the ascent of takeoff and the descent of landing – is around 35,000 feet. That’s nearly 7 miles up in the air. However, the number generally varies from about 33,000 feet to 42,000 feet.

When an object is dropped from an Aeroplane moving horizontally with constant velocity?

( R) : If a bomb is dropped from an aeroplane moving horizontally with constant velocity then horizontal component of velocity of the bomb remains constant and same as the velocity of the plane during the motion under gravity.

What happens if you drop something out of an airplane?

In the absence of significant drag, an object dropped from a moving plane will fall straight down relative to the plane, but the object will also move horizontally relative to the surface of the Earth.

How do you find the horizontal distance?

Horizontal distance can be expressed as x = V * t. Vertical distance from the ground is described by the formula y = – g * t² / 2, where g is the gravity acceleration and h is an elevation.

How do you find horizontal speed?

Divide Displacement by Time

Divide the horizontal displacement by time to find the horizontal velocity. In the example, Vx = 4 meters per second.

How long was the plane in level flight horizontal )?

The plane was in level flight for 80 minutes.

Hope this answer is helpful.

Which of the following forces act on the cart?

The forces on the cart include the forward force the horse exerts on the cart and the backward force due to friction at the ground, acting on the wheels. At rest, or at constant velocity, these two are equal in size, because the acceleration of the cart is zero.

What forces act on an object when it falls?

When an object is freely falling, the force due to gravity acts on it. Other than that, the air resistance acts on it which opposes the force due to gravity. Gravity being a stronger force prevails but, the net force is Gravity minus the air resistance.

How do the forces act on object?

For both stationary and moving objects with unchanging speed and direction, all the forces acting on the objects are in balance with each other, i.e. they all cancel each other out. Passive objects exert forces on objects that are exerting forces on them.

What distance do planes take off at?

That 35 feet is the safety margin determined by the FAA to ensure that the aircraft is climbing at an acceptable rate with one engine inoperative.

Why do planes fly at 35000 feet?

A balance between operating costs and fuel efficiency is achieved somewhere around 35,000 feet, which is why commercial airplanes usually fly at that altitude. Commercial airplanes can climb to 42,000 feet, but going beyond that can be precarious, as the air starts to become too thin for optimum flight of the airplane.

Why do planes cruise around 36000 ft?

The biggest reason for this altitude lies with fuel efficiency. The thin air creates less drag on the aircraft, which means the plane can use less fuel in order to maintain speed. Less wind resistance, more power, less effort, so to speak. Spending less on fuel is also great for airlines, for obvious reasons.

Aircraft Motion 2 – Answers

Aircraft Motion (Level 2) Answers You are the pilot of an airplane sent to drop supplies to victims of an accident that are stranded on a small island. The plane’s altitude is always 500 meters and your speed is 89.61 m/s.
  1. What amount of time will pass after the supply package has been released before it reaches the level of the island? (Hint: Use the distance equation in the y-direction to solve for the distance.) To begin, set the beginning distance to 500 meters, the acceleration to -9.8 meters per second2 (gravity, the negative sign denotes downward direction), the ultimate distance to zero meters, and the velocity to one millimeter per second (vi = zero millimeters per second). Step 2: Find the solution to the equation df = (1 / 2) at2 + vit + di at the time t. Is it possible that the time required for the supply package to descend will alter if the airplane’s speed changes? If the answer is a yes, the answer is a no.
  2. How far in advance of the island should the package be delivered in order for it to make landfall on the island (in terms of horizontal distance)? Step 1: During the fall, the package travels forward a horizontal distance of dx at a velocity of 89.61 meters per second and a time of 10.10 seconds. Step 2: dx = vt = 905.06 m is the solution. The package must be released by the airplane at a distance of at least 905.06 meters in front of the island.
  3. When the package reaches the level of the island, what is the horizontal speed of the package? It remains constant even when air resistance is taken into consideration. When the package reaches the level of the island, what is the vertical speed of the package? vx = 89.61 m/s
  4. what is the vertical speed of the package? Step 1: vi = 0 meters per second
  5. a = -9.8 meters per second2
  6. t = 10.10 seconds Step 2: For vf, solve the equation a = (vf – vi)/t using the variable a. In this case, vf = at + vi = -98.98 m/s (since vf = y)
  7. What will be the flight angle of the package in relation to the level of the island as it descends? Step 1: Create a diagram of the situation. Step 2: vy is equal to -98.98 m/s and vx is equal to 89.61 m/s. Step 3: tan Q= vy / vx
  8. Q = tan-1(vy / vx) = 47.84 o
  9. Q = tan-1(vy / vx) = 47.84 o For example, let us assume that victims on the island are able to grab supply shipments that have landed within 30 meters of the island. A 50-meter stretch of the island runs along the direction from which you are approaching it.
  10. For the airplane to land 30 meters in front of the island, how far in front of the island would the supply package have to be released before the plane could make the landing? dx(front) = distance from problem3 + 30 m = 935.06 m
  11. dx(back) = distance from problem3 + 30 m = 935.06 m
  12. It’s unclear how far in advance of the island the airplane would have to fly before releasing the supply package, in order for it to land 30 meters behind the island. dx(back) = distance from problem3 minus 50 meters minus 30 meters = 825.06 meters
  13. It is the time on your watch when you release the package and it lands 30 meters in front of the island that will be considered time1. When you release the package and it lands 30 meters in back of the island, the time on your watch will be the time on your watch. Prepare a calculation for the amount of time you have to drop the package successfully, i.e. time1- time2. Step 1: time1 = 0 seconds
  14. vx = 89.61 meters per second Solution of the equation dx = (vx)(time2) for the second time step. Also keep in mind that dx = dx(front) – dx(back) (back). 1 minute and 23 seconds = time2/vx = 1.23 sec As a result, time1 minus time2 equals 1.23 seconds.
  15. What could be done to slow down the package’s pace when it reaches its destination on Earth? Make use of a parachute

Delivering a package by air

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  • The thread was started by 05holtel on September 30, 2008.

Homework Statement

A food delivery is being delivered by a relief flight to a group of individuals who have become stuck on a very small island.Because the island is too small for the plane to land on, the only option for delivering the gift is to drop it off at its destination.At a height of 850 meters, the airplane flies horizontally at a constant speed of 290 miles per hour.

  1. Assume that the term ″island″ refers to the position at a distance D from the point at which the package is launched, as depicted in the figure, for all purposes.
  2. Ignore the fact that this place is elevated above sea level.
  3. In this example, we will assume that the acceleration due to gravity is g = 9.80 m/s2.
  4. a) How long will it take for a package to reach sea level after it has been expelled from a plane after it has been ejected?
  5. For simplicity, let us assume that the package has an initial horizontal velocity of 290 miles per hour, much like the plane.
  6. The package should be launched at a horizontal distance of D from the plane to the island, assuming that the package is to land directly on the island.
  1. The package strikes the ground with a speed of v f, which is measured in miles per hour.
  2. Equations x1=x0 + v0x(t1-t0) y1= y0 + v0y(t1-t0) -0.5(g)(t1-t0) -0.5(g)(t1-t0) -0.5(g)(t1-t0)

The Attempt at a Solution

A) 13.1 seconds b) 1710 milliseconds c) I’m not sure how to go about doing c. When it reaches the ground, I assumed the speed would be zero, but I’ve been informed that this is wrong.

Answers and Replies

At a height of 850 meters, the airplane flies horizontally at a constant speed of 290 miles per hour.Assume that the term ″island″ refers to the position at a distance D from the point at which the package is launched, as depicted in the figure, for all purposes.Ignore the fact that this place is elevated above sea level.

  1. Allow for the possibility that the acceleration due to gravity is 9.80 m/s2….
  2. The package strikes the ground with a speed of v f, which is measured in miles per hour.
  3. Equations are a type of equation.
  4. I’m not sure how to go about doing c.
  5. When it reaches the ground, I assumed the speed would be zero, but I’ve been informed that this is wrong.
  6. Hello, 05holtel!
  1. Hint: vf2 is equal to the sum of vhorizontal2 and vvertical2.
  2. vf2 is equal to the sum of vhorizontal2 and vvertical2.
  3. What does this imply?

Vf2 = 290mph2 + 0mph2 Vf = Square root of (2902) Vf = Square root of (2902) Vf=290mph vf2 is equal to the sum of vhorizontal2 and vvertical2.Does this suggest that Vf2 = 290mph2 + 0mph2 Vf = Square root of (2902) Vf = Square root of (2902) Vf=290mph umm…When it departs the plane, vvertical equals zero.Later on, you may use the standard formula to get vvertical 13.1s.

So, what does this imply: Vfy=Viy-g(Delta t) Vfy=0-4.38(13.1) Vfy=0-4.38(13.1) *** 9.8 meters per second = 4.38 miles per hour In this case, Vfy=751.65mph and Vfx=0mph.As a result, Vf is equal to 751.65 Is that what you’re saying?So, what does this imply: In the equation, Vfy=Viy-g(Delta t) Vfy=0-4.38(13.1)***9.8m/s = 4.38mph In this case, Vfy=751.65mph and Vfx=0mph.

As a result, Vf is equal to 751.65 Is that what you’re saying?No, Vfx is equal to 290mph.Vfy=Viy-g(t) is the correct formula.However, your g is incorrect.g does not equal 9.8 meters per second…It’s 9.8 meters per second…

  1. You must convert it to miles per second2.
  2. So, what does this imply: 9.8 meters per second squared = 0.006 089 miles per second squared Vfy=290mph-0.006089mile/s^2(13.1s) Vfy=290mph-0.7976m/s In the equation Vfy=290mph -1.178mph =288.8mph So, what does this imply: 9.8 meters per second squared = 0.006 089 miles per second squared Yes.
  3. Vfy=290mph-0.006089mile/s2(13.1s) Vfy=290mph-0.7976m/s Vfy=290mph-1.178mph =288.8mph Vfy=290mph-0.006089mile/s2(13.1s) Vfy=290mph-0.006089mile/s2(13.1s) Vfx is 290, and the speed is 0.07976m/s.

Where did the 1.178 come from, and what does it mean?The following are the results: Vfy=290mph-0.006089mile/s2 (13.1s) Vfy=290mph-0.7976mile/s Vfy=290mph-2871.36mph =?You’re missing the point completely.Is it possible that I converted it incorrectly?

The Plane and The Package

Consider the case of a plane traveling at a constant speed at a high altitude above the surface of the Earth.During the course of its trip, the plane loses sight of a package that was stored in its luggage compartment.In what direction will the package travel and where will it be in relation to the plane is unknown.

  1. In addition, how would you describe the motion of the package?
  2. A situation like this is seen in the animation below.
  3. It is possible to see the course taken by the plane and the package; in addition, the animation depicts the velocity components (horizontal and vertical), which are indicated by arrows.
  4. For the purposes of illustration, the package follows a parabolic route and always remains exactly below the plane, as shown in the animation above.
  5. During the descent of the package, it experiences vertical acceleration, which is to say, it experiences a change in vertical velocity.
  6. This vertical acceleration is related to the downward force of gravity acting on the package, which causes it to rise vertically.
  1. Alternatively, if the package’s motion could be approximated as projectile motion (i.e., if the impact of air resistance could be considered to be low), then there would be no horizontal acceleration seen.
  2. There would be a constant horizontal velocity in the absence of any horizontal forces acting on the object.
  3. This explains why the box would be dumped immediately beneath the plane from which it was delivered.

Many might argue that the package is being acted upon by a horizontal force because it is moving in a horizontal direction….This just isn’t true in this instance.The package’s horizontal motion is the consequence of its own inertia, which causes it to rotate.When the cargo was thrown from the plane, it was already moving in a horizontal direction.

A horizontal force will not cause the package to change its condition of horizontal motion unless it is operated upon by another horizontal force.An item in motion will continue to move at the same pace and in the same direction as it was before it started.(This is Newton’s first law.) Maintain constant awareness of the fact that forces do not create motion; rather, forces cause accelerations.

You may learn more about physical descriptions of motion by visiting the The Physics Classroom Tutorial website.There is much material on the following topics available on the website: Gravity is being accelerated.Gravity’s acceleration and the independence of mass are two important concepts in physics.Projectiles A projectile’s trajectory has certain characteristics.Newton’s First Law of Motion is a fundamental principle of motion.The concept of inertia and the state of motion

What Is the Altitude of a Plane in Flight?

The yoke is a control device that is used to adjust the height of a plane.(Image courtesy of Jupiterimages/Comstock/Getty Images.)…) Commercial jet jets fly quite high, mostly so that you can’t hurl rocks or day-old bread at them, as you might with a smaller plane.

  1. That’s not the true reason, to be honest.
  2. In contrast, the normal ″cruising altitude,″ which is the maximum height achieved during a flight and maintained between the climb of takeoff and the fall of landing, is around 35,000 feet.
  3. That’s approximately 7 miles above the surface of the earth.
  4. According to most sources, the elevation ranges between 33,000 and 42,000 feet.
  5. Private jets often fly at the upper end of the range, flying at roughly 41,000 feet, in order to travel the most direct path possible, above and beyond any commercial aircraft.
  6. Light aircraft are often restricted to altitudes below 10,000 feet.

Is It Arbitrary?

Nope.There are a variety of valid reasons why commercial planes fly above 35,000 feet or somewhat higher.When air is thin enough to reduce drag (and, as a result, enhance fuel economy and lower operating costs), yet there is still enough oxygen to power the engines, this is known as ″sweet spot″ flying.

  1. The troposphere, for those of you who weren’t paying attention in science class, is the layer of Earth’s atmosphere closest to the ground where the majority of the weather happens, and it reaches up to around 36,000 feet above the surface.
  2. In addition to reducing turbulence, flying at the top of this layer allows the plane to avoid flying through thunderstorms and other weather phenomena to a large extent.
  3. It also elevates planes above other flying objects and wildlife such as birds, swarms of locusts, propeller and single-engine planes, and helicopters, among other things.
  4. Furthermore, despite it may not seem like much, flying at a higher altitude provides pilots more time to correct problems or prepare for an emergency landing if something goes wrong.
  5. Light aircraft, on the other hand, are restricted to altitudes below 10,000 feet due to the lack of pressurized cabins.
  6. If they were to rise any higher, everyone on board – including the pilot – would be required to use oxygen masks for the duration of the journey.

Is It Scary?

Nope.Why do commercial aircraft fly at 35,000 feet or a bit higher than that?There are several valid reasons for this.

  1. This is the sweet spot when the air is thin enough to significantly minimize drag (and, as a result, boost fuel economy and lower operating costs), but yet contains enough oxygen to power the engines.
  2. As a side note, most weather happens in the troposphere – for those of you who were absent from science class – which is the lowest layer of the Earth’s atmosphere that stretches to around 36,000 feet above the surface of the Earth.
  3. In addition to minimizing turbulence, flying at the top of this layer allows the plane to avoid flying through thunderstorms and other weather phenomena to a large extent.
  4. Also included are birds, swarms of locust, propeller and single-engine planes, as well as helicopters, which are all placed above planes.
  5. And, though it may not seem like much, flying at a higher altitude provides pilots more time to troubleshoot problems or prepare for an emergency landing if something goes wrong on the ground.
  6. Light aircraft, on the other hand, are restricted to altitudes below 10,000 feet because to the lack of pressurized cabins in them.
  1. If they were to rise much higher, everyone on board – including the pilot – would be forced to use oxygen masks for the duration of the journey.

Is It Sickening?

Additionally, some people endure severe ear popping, motion sickness, or other bodily discomfort while on a flight, in addition to their dread of flying.Ear popping happens when the altitude and air pressure vary rapidly, like during takeoff and landing, which is especially true during the fast ascent and fall during takeoff and landing.For some airline travelers, clogged ears are not only uncomfortable, but they may be downright painful.

  1. In addition, a stuffy nose might exacerbate the situation even worse.
  2. Talking, swallowing, yawning, chewing gum, sucking on a hard candy are some of the most effective cures.
  3. Other options include utilizing an oxymetazoline nasal spray or wearing special earplugs manufactured specifically for this purpose, which are available at most drug stores.
  4. Direct flights are also better than connecting flights for persons who suffer from severe ear discomfort caused by popping ears; being in the air for a longer period of time does not exacerbate clogging, but making many journeys up and down from the ultimate cruising altitude does.
  5. When traveling by plane, those who suffer from motion sickness should abstain from drinking alcohol and eating heavy or greasy meals in the 24 hours before to taking off and while on board the plane.
  6. Make sure you pack a simple, light lunch that is TSA-approved in your carry-on, remain well hydrated, and avoid reading on the plane.
  1. When that sensation of discomfort or illness begins to set in, look at a fixed spot in the cabin for support.
  2. When traveling by coach, seek a seat towards the front of the plane or near the base of the wings to reduce exposure to the movements that might cause motion sickness in certain people..
  3. Open the upper air vents all the way and direct the air directly onto your face to keep you cool.

Some individuals find that ginger relieves an upset stomach, and there’s an acupressure method that includes pushing down on the middle of the wrist 2 inches below its crease and varying the amount of pressure until you feel comfort; then maintain the position for a few minutes to see whether it works for you.Another alternative is to self-medicate with the antihistamine dimenhydrinate, but keep in mind that this medication might induce substantial sleepiness in some people.ReferencesResources Biography of the Author The author, travel, food and drink, and lifestyle writer Eric Mohrman now resides in Orlando, Florida, which is one of the world’s most popular tourist and business travel destinations.This affords lots of opportunities for him to share travel advice and local insights with other tourists, and he particularly appreciates highlighting lesser-known sites and experiences that are sometimes overshadowed by the region’s major theme parks and attractions.

He has written travel articles for a variety of publications, including Visit Florida, MapQuest, Dollar Stretcher Magazine, USA Today 10Best, Agent Magazine, The 863 Magazine, Working Mother, Downtown Orlando Community Paper, and others.He is a graduate of the University of Florida.

Horizontal Projectile Motion Calculator

  • In fact, horizontal projectile motion equations are a specific example of generic formulae, as we previously discussed. Due to the fact that the launch is parallel to the ground (and thus the angle is equal to 0°), we do not need to mention the angle of launch. As a result, we only have one component of initial velocity – Vx = V – and no component of initial velocity – Vy = 0. We will assume that the starting location is at the beginning of the journey. The following is how the equations of motion may be written: Distance The horizontal distance may be calculated using the formula x = V * t.
  • An elevation is represented by the vertical distance between it and the ground, which can be calculated using the formula: y =– g*t2/2, where g is gravity acceleration and h is elevation.
  • Velocity Horizontal velocity is equal to V
  • vertical velocity is equal to V
  • In the case of vertical velocity, –g * t can be used to indicate it.
  • Acceleration Horizontal acceleration is equal to zero
  • vertical acceleration is equal to one.
  • -g denotes negative vertical acceleration (since only gravity exerts an effect on the projectile)

According to the equations of horizontal projectile motion, they are as follows: A trajectory’s equation is defined as

Combining x = V*t and y =–g*t2/2 will allow us to eliminate the term ″t″ from the equations. The following is the trajectory: y =– g * (x / V)2 / 2 = (- g * x2) / (2 * V2) = (- g * x2) / (2 * V2) The flight’s departure time

In order to determine the period of flight of the projectile, we must first determine when the projectile makes contact with the ground.Specifically, it occurs when the y coordinate equals the h coordinate, which is represented by the equation: g * t2 / 2 = h.We may deduce from this equation that the time spent in flight is equal to: t = ((2 * h / g)) is the time in seconds.

  1. The projectile’s range is given in the table below.

The projectile’s range is defined as the entire horizontal distance covered by the projectile throughout its flight period.The equation r = V * t = v * (2 * h / g) may be written down as r = V * t = v * (2 * h / g).In this case, we won’t compute the maximum height since we don’t have an initial vertical velocity component – and since we don’t have an initial vertical velocity component, the maximal height is equal to where we started.

  1. We did not account for the air resistance acting on the projectile in any of our calculations, and as a result, the total of kinetic and potential energy is preserved.
  2. More information regarding the latter may be found in our potential energy calculator.

How to Calculate Horizontal Velocity

When working on velocity issues in physics, you divide the motion into two parts: the vertical component and the horizontal component.It is necessary to employ vertical velocity when solving issues that include an angle of trajectory.When things are travelling in a horizontal direction, the horizontal velocity of the item becomes crucial.

  1. Because the horizontal and vertical components are completely independent of one another, any mathematical solution will regard them as separate entities in its calculations.
  2. In general, horizontal velocity is defined as horizontal displacement divided by time, as in miles per hour or meters per second, respectively.
  3. Quite simply, displacement is the distance traveled by an item relative to its initial location.

TL;DR (Too Long; Didn’t Read)

Whenever you are dealing with motion in physics issues, you should regard horizontal and vertical velocities as two different, independent entities.

Identifying the Horizontal Velocity

It is the horizontal velocity of a motion issue that is concerned with motion in the x direction; that is, movement from side to side, rather than movement up and down. As an example, gravity only affects motion in the vertical direction and does not have any effect on horizontal motion at all. It is caused by forces acting along the x-axis that cause horizontal motion.

Tips for Recognizing Horizontal Velocity

It takes a lot of work to be able to discern the horizontal velocity component in a motion issue.A ball being thrown forward, a cannon launching a projectile, or a car driving along a highway are all examples of situations in which horizontal velocity exists.A rock dropped straight down into a well, on the other hand, has no horizontal velocity, just vertical velocity, and hence cannot be moved.

  1. Some objects, such as a cannonball fired at an angle, will have a mix of horizontal and vertical velocity.
  2. For example, a cannonball fired at an angle will travel both horizontally and vertically.
  3. However, even though gravity only operates in the vertical direction, you may have an indirect horizontal velocity component, such as when an item rolls down a ramp, when gravity is acting in the vertical direction.

Writing the Horizontal Component

In the case of a general velocity problem, you can simply write an equation with the letter ″V″ representing velocity, such as:V=a times t.It is necessary to distinguish between horizontal and vertical velocity in order to write a motion equation that treats them separately.To do so, you must use different symbols for horizontal and vertical velocity, respectively: Vx and Vy.

  1. If the problem specifies both horizontal and vertical velocities, you must write them as two separate equations, such as these: If the problem specifies both horizontal and vertical velocities, write them as two separate equations, such as these: TextV y=-9.8 times fractextV x=25 times fractextV y=-9.8 times fractext

Solving a Horizontal Velocity Problem

Vertical velocity is expressed as V x=frac, where Vx is the horizontal velocity of the axis of rotation. As an illustration, V x=frac=4 text

Divide Displacement by Time

Vertical velocity is expressed as V x=frac, where Vx is the horizontal velocity of the arrow. As an illustration, V x=frac=4text.

Calculating Negative Velocity

Vertical velocity is expressed as V x=frac, where Vx is the horizontal velocity. As an illustration, V x=frac=4text

The Horse and Cart Problem.

Write the horizontal velocity issue as V x=frac, where Vx represents the horizontal velocity. For example: V x=frac=4text


  • The paragraph contains a number of errors, including ambiguity concerning whose body the forces are acting on. Failure to separate the system into subsystems for the purpose of analysis.
  • Newton’s third law states that various bodies are subjected to different forces. Forces that contribute to the net force on a body act on that body alone, rather than on other bodies.
  • A failure to define the system that is speeding up.

It is generally better to study the cart and the horse individually in order to adequately understand this situation.The following is the procedure taught to students in introductory physics courses: first, identify and isolate the body to which you intend to apply Newton’s second law, then identify all forces acting on that body and only on that body, add them (as vectors) to get the net force, and finally, use Fnet = ma to solve the problem.

Forces on the cart, no acceleration. Forces on the horse, no acceleration.
  1. In these diagrams, an oval or a circle has been utilized to denote the boundary of the subsystem under consideration.
  2. On the cart, there are two forces at work: the forward force exerted by the horse on the cart, and the backward force caused by friction at the ground, which acts on the wheels.
  3. Because the cart’s acceleration is zero while it is at rest or moving at a steady speed, these two objects are of identical size.
  4. The horse is subjected to a variety of pressures, including the backward force exerted by the cart and the forward force exerted by the earth on its hooves.

Because the horse’s acceleration is zero either at rest or at a constant velocity, these two objects are of same size when at rest or constant motion.As a result, A = -B.Newton’s third law states that the force exerted by the horse on the cart is equal in magnitude and opposite in direction to the force exerted by the cart on the horse.Because B = -C, and this is true whether or not anything is accelerating, we may write B = -C as a pair of forces.Because the horse is not speeding, C = -D is established by Newton’s second law, and we can see that all of the forces depicted in the picture are of equal size.As the horse exerts more effort, both the horse and the cart begin to move, increasing from zero to a certain speed.

  1. It is necessary for the net forward force on the horse to be larger than the net backward force on the horse during that acceleration to be successful.
  2. A further requirement is that the net forward force applied to the cart must be larger than the net backward force applied to it.
  3. This is a quote from Newton’s second law of motion.
Forces on the cart when accelerating. Forces on the horse when accelerating.
  1. So, what has changed since then?
  2. The friction between the horse’s feet and the cart’s wheels is now larger than the friction between the horse’s feet and the cart’s wheels.
  3. The friction on the cart’s wheels is known as rolling resistance, and it is mostly determined by the amount of the cart’s load, rather than the cart’s speed, when the cart is moving.
  4. As a result, not much has changed.

However, because it is gaining speed, the power that the horse puts on the cart has grown as well.Newton’s third law states that the force of the cart on the horse has risen by the same amount as the force of the cart on the horse.However, because the horse is also speeding, the friction between the ground and the horse’s hooves must be greater than the force exerted by the cart on the horse.There is static friction between the hoof and the ground, not sliding or rolling friction, and it may be increased as necessary (up to a limit, when slipping might occur, as on a slippery mud surface or loose gravel).Because of Newton’s third rule, we still have B = -C when accelerating, but D > C and B > A, resulting in D > A.Vector sizes are denoted in lower case without consideration to the sign or direction of the vector.

  1. In addition, D-C = Ma, where M is the mass of the horse, and B-A = Ma, where m is the mass of the cart, are true.
  2. You may even come to the conclusion that D-A = (M+m)a by assuming the system to be the horse and cart working in conjunction with one another.
  3. It is important to note that the net force acting on the entire system is D+C+B+A = D+A, because B = -C according to Newton’s third rule.
  4. Because A is less than D and in the opposite direction of D, the net force acting on the system has the size D-A and is in the forward direction.

B and C are regarded to be an internal reaction force pair, and as a result, they are always equal and opposing to one another, and they always add vectorially to one another.As a result, they are typically neglected when calculating the sum of forces acting on the entire system.We have shown that certain information about a system can only be recovered by breaking it down into component pieces, and that these forces inherent to the entire system cannot be ignored since one of them represents an external force for a subsystem.It all starts with the horse, as you might expect.The horse positions its feet in such a way that the angle of the force exerted by its hooves on the ground is changed, so increasing the backward force exerted by the horse’s hooves on the ground.

Newton’s third law states that the backward reaction force of the ground on the horse’s hooves rises as a result of this.As a result, the strain on the harness grows, and the force on the cart increases as well.When the horse reaches the required forward speed, he reduces his effort, and at constant speed, all of the horizontal forces we discussed before become equal in size again, and there is no additional acceleration.Please keep in mind that the action response pair B and C are depicted in the diagrams without being specific about where they act.In this type of puzzle, the principal components are connected by a rope or light rod to form a chain.In this instance, by means of a harness and a wagon tree.

  1. When compared to the horse and wagon, their mass is small, and their mass is thus neglected (treated as if it were zero).
  2. We should approach this as a three-body system, and we should construct equations for the third body, you could be concerned (the harness and wagon tree).
  3. However, because they are of tiny mass, they have little influence on the forces at work in the problem.
  4. This is because if just two forces are operating on a zero mass body, they must amount to zero, and as a result, they must be equal in magnitude and opposite in direction, as shown in the diagram.
  5. A nearly zero-mass body is subjected to the action of just two forces, both of which are nearly equal in magnitude and opposing in direction.

This is a logical conclusion to be drawn from Newton’s second law.Students become confused when textbooks assume this simplification in situations of this nature without explicitly stating or defending it, which is the case far too often.It was only necessary to evaluate the horizontal force components on the bodies in order to answer this question.

There is no contribution to the horizontal motion from the vertical force components, which include downward pressures owing to gravity as well as upward forces from the road acting on the cart and horse.Furthermore, because nothing is experiencing rotational acceleration, we don’t need to worry about torques.What is the driving force behind the horse’s forward motion?What you’re feeling is the ground’s gravitational pull.When the horse pushes backward on the ground, the ground pushes forward with the same amount of force as the horse.

Only when the horse pushes back against the ground with a force larger than the opposing force of the cart will the animal begin to speed.Furthermore, because the cart is attached to the horse, it must accelerate in tandem with the horse.How can anything move if every force has a response force of equal size and oppositely directed opposite to it?This is due to the fact that action and reaction forces operate on bodies in different ways.

Follow-up question

We’ve said before that since work is force × distance, so a force does no work on a body unless the body moves. But the force of the ground on each hoof does not involve motion of either the ground or the hoof. So how can the ground do any work on the horse? Answer. It doesn’t. The horse does all the work through motions of the muscles in its body. The ground provides an anchor for its hooves at each step. The ground (or pavement) isn’t doing any of that work. Sometimes we forget this as we do problems. Suppose you are on roller skates and you push yourself away from a solid wall. We often calculate the work done in that process by multiplying the force the wall exerts on you by the distance you move as you push. (Actually an integral is required if that force is not constant.) But that work is still supplied by your muscles, not the wall, and the energy gained comes from stored energy from the food you’ve eaten. Revised June 26, 2011. Input and suggestions are welcome at the address shown to the right. When commenting on a specific document, please reference it by name or content. Return to Donald Simanek’s page.

Forces on stationary objects

The following are some examples of how this focus notion is explored:

Contrasting student and scientific views

Student everyday experiences

  1. For this reason, students frequently assume that immobile or passive objects are not subjected to any pressures at all, because they equate pushes and pulls with moving or active objects.
  2. In the focal idea Pushes and pulls, this concept is also examined in further detail.
  3. For example, pupils believe that an active item such as a human hand may feel and generate forces, but a passive object such as a book laying at rest on a table is completely devoid of all forces.
  4. While this viewpoint may have been widespread among students throughout the middle school years, some students at the senior secondary level may still hold this viewpoint.

Gunstone and Mitchell (1998) conducted research, as did Gunstone and Watts (1998).(1985) Objects breaking or bending when large forces are applied to them is something that students are familiar with, but they are less likely to recognize that forces are there when there is no visible change or distortion when the force operates.An apple lying in a fruit dish, on the other hand, is not considered to involve forces at work.For example, a chair falling under the weight force of a person or a trampoline stretching are both considered to involve forces at work.Clement’s investigation (1987)

Scientific view

  1. If all of the forces acting on the things are in balance with one another, i.e., if all of the forces acting on the objects have the same speed and direction, then the objects are said to be stationary.
  2. Passive objects exert forces on things that are exerting forces on them, and passive objects exert forces on themselves.
  3. Because of its downward weight force, a heavy bag laying on a bed exerts downward pressure on the bed, which then exerts a balancing force by pushing up on the bag as it is squished under its own weight and weight force.
  4. When a force is applied to an item, it undergoes some squashing (compression) or stretching, and its form is altered accordingly.

Compression occurs in the case of strong materials such as concrete or steel table frames, although it is often very tiny and not noticeable until the applied force is quite significant.

Critical teaching ideas

  • There are forces acting on stationary things.
  • When a force is applied to an item, it will experience some squashing or stretching.
  • A push between two objects results in the first thing squashing the second object, which then pushes back.
  • In the Concept Development Maps: Laws of Motion, you may learn about the links between thoughts about forces and how they are related. A squashing effect occurs when one thing pushes against another and the surface of the second object gets squished. Because it has been squished, the second object acts as a counterweight to the first. If a person sits on a chair that bends as a result of the bend, the chair pushes back up on the person who is sitting in it. Loughran, Berry, and Mulhall conducted research (2006) Examples or circumstances for discussion that will assist students in identifying the forces operating on passive items such as tables, simple bridges, beds and chairs should be chosen to aid in the identification of the forces acting on passive objects. Make an argument supporting the idea that certain forces occur as a result of things being stretched or squished. Develop the concepts that: when something is crushed or stretched, it changes form and pushes or pulls on the object that is squashing or stretching it
  • when something is squashed or stretched, it pushes or pulls on the thing that is squashing or stretching it
  • Some items are more readily squished or stretched than others
  • Generally speaking, the more something is squished or stretched, the more forcefully it pushes or pulls on the item that is crushing or stretching it.

Gunstone and Mitchell conducted research (1998)

Teaching activities

Open up discussion via a shared experience

When items are being stretched or squished by huge and little pressures, students should pay attention and participate in discussions about what they are seeing. If they can feel the forces being exerted by the items that have been stretched or squished, it will be easier for them to verify that the forces involved are indeed present.

Challenge some existing ideas

  • Example: A heavy object may be put on top of an inflated balloon or a piece of soft foam to provide support. In order to replicate what they have witnessed, students must first take an object and replace it with their finger, then feel the balloon/foam pushing back up as they attempt to make the balloon resemble a comparable form.
  • A metre ruler is bent by placing several heavy books in the centre of it and supporting it at both ends. This is an example of POE (Predict-Observe-Explain). In order to prevent the ruler from rebounding as the books are progressively taken off it, students lay their finger beside the books on the ruler. This allows them to provide the necessary push to keep the ruler bent. If the students move the supports closer together and repeat the sequence of activities previously described, they might be asked to guess what will happen. They then observe and describe what they have observed.
  • Using the POE method (Predict-Observe-Explain), do the same practice as before, but with fewer books, or lay a hefty weight/brick on the top of a sealed plastic soft drink container. Take note of the squashing. After that, ask the students to guess what would happen if they lay a weight on top of the bottle while the lid is still on.
  • It is necessary to suspend an object from a rubber band in order for it to stretch. It is necessary for students to feel the pull of the rubber band at the same extent as they did before the weight was removed.

Challenge some existing ideas

  1. It is very uncommon for students to be astonished to learn that bridges are meant to bend when automobiles pass over them, and that big structures are built to wobble in severe winds or during earthquakes.
  2. Students should be given the opportunity to explore how they would go about gauging if a tall structure or a freeway was moving during a strong wind.
  3. The Millau Bridge in France is an excellent case study to look into further.
  4. The bridge is the world’s highest structure.

Making the Modern World: Bridges is an excellent website to learn about the design of bridges.

All About Runway Distance Requirements – Cal Aero Blog

  1. Being a safe pilot includes taking precautions to ensure that any runway you intend to utilize has sufficient distance to allow for safe operation during both takeoff and landing.
  2. This problem will be addressed via a robust student pilot program.
  3. Because dealing with runway distance is a subject that goes much beyond basic memorization, it is a crucial aspect of ensuring safe takeoffs and landings.

Planning and Wind

  1. The distance between takeoff and landing is often shorter: Just because you can take off from a runway does not imply that you can return to it to land on it.
  2. Weather conditions like as snow and ice increase the distance required for takeoff and landing and frequently disqualify a runway from consideration for operation.
  3. In certain circumstances, you must arrange to return to the field using a different runway, and if the airport from which you are departing has only one runway, you must pick another airport that is close to your destination.
  4. This is referred to as a takeoff alternative.

It’s similar to a landing alternative, however it’s used for departure planning purposes instead of landing planning.A significant factor in determining both takeoff and landing distances is the effect of wind.As a result, wind speed and direction are the first items shown on an airport’s hourly weather report, known as the METAR.Takeoff and landing must be done into the wind or at the at least with a crosswind for pilots of any aircraft type, whether it be an airplane, a helicopter, or something else.It is possible to fly with a headwind because it minimizes takeoff ground roll, allowing for shorter runway lengths and less wear on the tires.Additionally, a headwind reduces the groundspeed during touchdown, lowering the distance necessary to halt and reducing tire wear throughout the landing process.

  1. However, taking off and landing with a tailwind is not recommended.
  2. In fact, most aircraft have a declared tailwind restriction to protect the aircraft’s performance.
  3. When that restriction is exceeded, damage to the landing gear, excessive tire wear, and even the inability to land or takeoff within the allocated runway distance might result as a result.

Airplane Structure and Runway Distance

  1. During the certification process for a new aircraft, designers and engineers calculate the distances that they believe the aircraft would require for different flap settings, weights, and runway conditions, among other things.
  2. An unpaved runway with regular flap settings and light weight will require significantly more distance than an unpaved runway with abnormal flap settings and heavy weight on a dry, level, paved runway with normal flap settings and low weight.
  3. Once the prototype of the new aircraft has been created, test pilots will fly it through every possible combination of runway conditions, flap settings, and weights to ensure that the computed distances are accurate and reliable.
  4. There are several excellent films of aircraft certification tests available online, including landings and brake tests over massive pools of water.

As soon as the actual distances are known, a safety margin is applied to those figures.The precise margin is determined by the portion of the certification requirements the aircraft is constructed under at the time of construction.

Runway Distance and Calculation

  1. The distance necessary to accelerate to rotation speed, refuse the takeoff, apply the brakes, and come to a complete stop is an essential metric to calculate first.
  2. This is known as the accelerate-stop distance.
  3. This distance is not required for certification for the most majority of single-engine aircraft; nonetheless, it is a good idea to calculate it for safety reasons.
  4. Because the accelerate-stop distance for tiny, single-engine airplanes is rarely published, a decent rule of thumb is to twice the computed takeoff distance.

Attempting a takeoff is not recommended if the runway provided does not exceed the distance required to accelerate and halt.Takeoff is forbidden for airline operations and other multi-engine turbojet aircraft if the distance between the accelerate-stop point and the runway length exceeds the runway length.Next, with a multi-engine aircraft, you must figure out how far you must travel to attain rotation speed, have an engine fail, yet still complete the takeoff since you have passed V1 speed and have climbed to 35 feet over runway altitude.That 35 feet is the safety buffer established by the Federal Aviation Administration to guarantee that the airplane continues to climb at an appropriate pace even if one of its engines is not operational.Generally speaking, for single-engine airplanes, the distance required to accelerate to takeoff speed, climb at Vx (best angle of climb), and clear at least a 50-ft high obstacle—off the departure end of the runway—is published in the Pilot’s Operating Handbook and must be taken into consideration when flying on short runways.When approaching a landing strip, a pilot must calculate the distance necessary to pass the threshold, touch down, and come to a complete stop on the runway.

  1. These lengths are likewise published in the POH and contain adjustments for factors such as wind, weight, flap settings, and surface conditions.
  2. For single-engine airplanes, a short-field landing distance is given, which is the inverse of the short-field takeoff distance stated.
  3. A 50-foot barrier must be crossed, the aircraft must land, apply maximum brakes and come to a complete halt in that distance.
  4. In the case of multi-engine airplanes, different landing distances are reported for two-engine landings vs single-engine landings in order to account for the difference in approach speeds between engines.

When using a turbojet, the landing distance includes touching down at the one-thousand-foot marker, and various lengths are computed depending on whether or not the thrust reverser is being utilized.In the aviation industry, a 60 percent safety margin is included to all landing calculations in order to provide a large safety margin for passengers.

Runway Condition Codes

  1. Runway condition codes, often known as RCCs, are a recent discovery in the field of landing distance computations.
  2. RCC is an effort to provide a numerical value to the condition of a runway surface, a number that pilots may use as a multiplier to their ″un-factored″ landing distance, which is the distance determined by the engineers and published in the aircraft’s Pilot’s Operating Handbook (POH).
  3. RCC takes the role of braking action reports, which was a very subjective mechanism for determining how slippery a runway was before the use of RCC.
  4. A common method of determining braking action was for an airport operations agent to drive out on the runway in a truck, accelerate to a specific speed, then slam on the brakes and time how long it took for the truck to come to a complete stop.

Braking action was classified as good, fair, bad, or non-existent based on the time it took or how the driver felt when decelerating.The challenge was determining the difference between good and fair or fair and terrible braking performance, and various pilots would have varying perceptions of how that braking motion would effect their airplane’s stopping capabilities.A data matrix containing information on precipitation type and depth is used to calculate RCCs.RCCs are presented as levels 0 through 6, with zero representing the former ″nil″ condition and 6 representing clear and dry conditions.In addition to his present position as President of California Aeronautical University, Mr.Matthew A.

  1. Johnston has more than 23 years of experience in a variety of educational capacities.
  2. The University Aviation Association (UAA), the Regional Airline Association (RAA), the AOPA, the National Basketball Association of America, and the EAA’s Young Eagles program are just a few of the organizations in which he retains membership and participates as a supportive member or volunteer.
  3. It gives him great pleasure to work in conjunction with airlines and aviation enterprises, as well as with individual aviation experts, in order to establish California Aeronautical University as a leader in the education of aerospace and aviation professionals.

Why Do Airplanes Need To Fly So High?

  1. Approximately 35,000 feet is the altitude at which a satisfactory balance between operational expenses and fuel efficiency can be reached, which is why commercial airplanes often fly at that level.
  2. The majority of commercial airplanes fly at a height of over 35,000 feet, or around 6.62 miles (10,600 meters) above the ground!
  3. For those who believe that the top of a skyscraper is impressive, flying more than 6 miles above the earth is truly amazing!
  4. After all, what’s wrong with flying a few hundred meters above the surface of the earth, so long as the plane clears all ground structures such as communication towers and city skylines?

The first thing to note is that the altitude at which most planes travel is not some arbitrary quantity determined by chance.Planes travel at specific altitudes in the sky for a variety of very valid reasons.Examine the scientific basis for why planes soar so high in the air.

Air resistance and fuel efficiency

  1. Air resistance is one of the primary reasons commercial airplanes soar to such great heights.
  2. As you can see, the higher you fly above the earth’s surface, the thinner the atmosphere gets, and the less resistance there is to the plane’s movement.
  3. The higher a commercial airplane flies, the less air resistance it encounters.
  4. (Photo courtesy of Pxhere.) Actually, it’s quite simple: the greater the number of air molecules that the plane must fly through, the greater the amount of energy it will require, and subsequently, the greater the amount of fuel it will require, and hence, the higher the running expenses.

Additionally, it is extremely cold up there, with temperatures hovering about -55°C, which helps to improve the efficiency of the jet engines.

Cruising altitude: a sweet spot

  1. Because of the decreased resistance at higher altitudes, commercial aircraft are able to maintain their forward motion with minimum fuel use.
  2. Commercial airplanes normally fly between 32,000 feet and 38,000 feet in height, with the sweet spot being roughly 35,000 feet, which is referred to as cruising altitude in the aviation industry.
  3. Approximately 35,000 feet is the altitude at which a satisfactory balance between operational expenses and fuel efficiency can be reached, which is why commercial airplanes often fly at that level.
  4. Commercial airplanes can fly up to 42,000 feet in altitude, but flying higher than that can be dangerous since the air becomes too thin for the airplane to fly at its best performance.

It is also becoming increasingly difficult for oxygen to reach the cells, making it impossible for fuel engine systems to function properly.Flight 3701 of the Pinnacle airline went down in 2004 because the pilots were overconfident and flew their aircraft at heights higher than the aircraft manufacturer recommended.After passing 41,000 feet in altitude, the pilots lost control of the plane, which crashed onto a roadway near the city of Jefferson City.During the crash of Pinnacle airline flight 3701 in 2004, this photo was taken by Aeroprints/Wikimedia Commons.The fact that commercial aircraft travel above 6.6 miles above sea level also serves as a key factor in their decision to do so.At that altitude, they have more’stable’ air and are less likely to be affected by clouds or other weather-related phenomena (e.g., thunderstorms).

  1. Although planes may still fly through clouds and storms, they will endure a great deal of turbulence as a result of the maneuver.
  2. As a result, passengers may experience discomfort and, in some cases, terror on the aircraft.
  3. To their advantage, pilots are well-versed in dealing with such circumstances, and they are required to report any such incidents to the air traffic control tower for additional assistance and advice.
  4. As an airline firm, you don’t want turbulence on your flights; instead, you want to provide your passengers with the smoothest trip possible.

Clearing obstacles

  1. This one is a little self-explanatory.
  2. In order to fly a plane, a pilot must avoid flying through towers, buildings, and other ground objects at high speeds.
  3. That is absolutely hazardous, not to mention impractical, regardless of how awesome it may seem to some of you.
  4. Because topography is defined by the sea level, some terrain may be significantly higher above sea level than the runway/airstrip, and vice versa.

In order to avoid any type of ground constructions, planes must first ‘climb’ to an acceptable height before taking off.Most aviation authorities throughout the globe prohibit commercial airplanes from flying below 1,000 feet above ground level unless there is an emergency.This is done for reasons of safety and the environment, such as noise.If you’ve ever been near an airport, you’re probably aware that commercial jets are quite loud and distracting.At roughly 35,000 feet, airplanes, on the other hand, are on the cusp between the troposphere and the stratosphere, where they may fly freely.Because this environment is almost completely devoid of birds, insects, and microorganisms, the likelihood of a birdstrike occurring there is very negligible.

Miracle on the Hudson

  1. Several cases of bird strikes on airplanes have been documented, but the one that garnered the most attention was the incident involving US Airways Flight 1549.
  2. In the early hours of January 15, 2009, an airliner (an Airbus 320) was hit by a swarm of birds, causing the plane’s engines to fail on both of its engines.
  3. Because the pilot was concerned that the jet would not be able to make it to the airport or to a safe location nearby, he had no choice but to allow the plane to plummet into the Hudson River and sink.
  4. It was a miracle on the Hudson when passengers were able to escape out of a plane that had crashed into the Hudson River.

(Photo courtesy of Wikipedia.) By some miracle, the plane did not sink after touching down in the river and instead managed to stay afloat.This gave the rescue crews enough time to complete their work and remove all of the passengers.The Miracle on the Hudson is a well-known name given to this occurrence.You may find out more about it by watching the video below.You’ll discover exactly what occurs when birds collide with an airplane and how such events may be avoided by flying at a sufficiently high altitude!You should watch the following video: As a precautionary measure,

  1. Consider the following scenario: you’re flying a commercial airplane about a mile above the earth when something goes wrong and you crash.
  2. The plane begins to tumble to the ground.
  3. You are aware that the problem that is causing the plane to plummet rapidly may be resolved; but, the plane is descending too quickly, and you simply do not have enough time to correct the problem before it crashes.
  4. You might say to yourself, ″If only I had more time…″ at that point.

Being in the air gives you that extra time to think, calculate, and put into action what you need to do to ensure the safety of your passengers.

What about private jets and helicopters?

  1. Up until this point, we’ve spoken about commercial flights reaching tremendous heights and reaping the benefits of increased altitude, such as less air resistance and bird irritation, but what about private planes and helicopters?
  2. When it comes to private jets, the engines are typically powered by a single piston piston-driven engine.
  3. This engine operates in a manner similar to that of an automobile.
  4. Its purpose is to provide more power to the plane for a shorter distance.

This engine does not have the ability to propel a plane into the upper echelons of commercial aviation flight.According to aviation experts, the planes we can hire or buy for private use are often not capable of flying

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