# Why is my answer wrong? Projectile Motion

• reyrey389
In summary: I get 26.9 m/s.In summary, an airplane with a speed of 70.6 m/s and an angle of 40.0 ° with respect to the horizontal releases a package when its altitude is 814 m. The distance along the ground from the point of release to where the package hits the earth can be calculated by finding the time of flight and multiplying it by the horizontal component of the velocity. The angle of the velocity vector of the package just before impact relative to the ground can be found by using the final velocity in the y direction and inverse tangent. In another problem, a car must have a minimum speed of 26.9 m/s to successfully complete a stunt where it drives up and off a
reyrey389

## Homework Statement

An airplane with a speed of 70.6 m/s is climbing upward at an angle of 40.0 ° with respect to the horizontal. When the plane's altitude is 814 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

## Homework Equations

70.6 * cos(40) = 54.1 velocity initial in the x direction = velocity final
70.6 * sin(40) = 45.4 velocity initial in the y direction

Displacement(Y) = -814 meters (initial to final position)

## The Attempt at a Solution

I already obtained the answer for part A. What I did was solve for t using

-814 = 45.4*t - 4.9t^2

t = 18.3 = time of flight

so for part a) the horizontal distance = 18.3 * 54.1 = 990 meters

now for part b) I have the final velocity in the x (54.1), I need the final velocity in the y, and I can use inverse tangent to find the angle once v_final(y direction) is found.

To find this value I used v_y = v_initial(y) - g*t using 45.4 for v_initial(y) and 18.3 for time
which gives v_y = -133.94 and so theta = inverse tangent ( 133.94/54.1) = 68 degrees from the ground, however this is wrong .

I'm assuming my value for the final velocity in the y direction is wrong...

I haven't checked your arithmetic, but your methodology looks right. They don't really specify how the angle is measured with respect to the ground. Is it possible that they are looking for the supplement of your angle?

Chet

Chestermiller said:
I haven't checked your arithmetic, but your methodology looks right. They don't really specify how the angle is measured with respect to the ground. Is it possible that they are looking for the supplement of your angle?

Chet

I figured out why I was wrong, not sure how to delete posts, but while this is up I also had another question on a problem a bit more involved.

In a stunt being filmed for a movie, a sports car overtakes a truck towing a ramp, drives up and off the ramp, soars into the air, and then lands on top of a flat trailer towed by a second truck. The tops of the ramp and the flat trailer are the same height above the road, and the ramp is inclined 16° above the horizontal. Both trucks are driving at a constant speed of 10 m/s, and the trailer is 18 m from the end of the ramp. Neglect air resistance, and assume that the ramp changes the direction, but not the magnitude, of the car's initial velocity. What is the minimum speed the car must have, relative to the road, as it starts up the ramp?Since the truck in front is traveling at 0 m/s relative to the truck behind it, I used
v0(car) *cos*(theta) * t = 18

and since the displacement in the y direction is zero
0 = v0(car) *sin*(theta) * t - 4.9 t ^2
and thus t = [vo*sin*(theta)]/4.9 Is this a wrong setup ?...

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Looks OK.

Your answer for the final velocity in the y direction is incorrect. This is because the initial velocity in the y direction should be positive, as the package is initially moving upward. Therefore, the equation should be v_y = v_initial(y) + g*t. This will give you a positive value for the final velocity in the y direction, and your calculation for the angle will then be correct. Additionally, it is important to double check your units when solving equations - in this case, you should use meters per second for velocity and meters for displacement.

## 1. Why is my answer for projectile motion wrong?

There could be several reasons why your answer for projectile motion is wrong. One possible reason is that you made a mistake in your calculations or equations. Another reason could be that you did not properly consider all the variables involved in the motion. It is important to double check your work and make sure you have included all the necessary factors.

## 2. How can I improve my understanding of projectile motion?

To improve your understanding of projectile motion, it is important to review the fundamental principles and equations involved. Practice solving different types of problems and make sure you understand the concepts behind each step. You can also try visualizing the motion using diagrams or simulations to help solidify your understanding.

## 3. What are common mistakes when solving projectile motion problems?

Some common mistakes when solving projectile motion problems include not properly setting up the initial conditions, using incorrect equations or formulas, and not considering all the factors that may affect the motion. It is important to carefully read and analyze the problem before attempting to solve it, and to double check your work for any errors.

## 4. How does air resistance affect projectile motion?

Air resistance, also known as drag, can have a significant impact on projectile motion. As an object moves through the air, it experiences a force in the opposite direction of its motion due to air resistance. This force can cause the object to slow down and change its trajectory, leading to a different path and final destination than what would be predicted without air resistance.

## 5. Can real-life projectile motion be perfectly modeled using equations?

No, real-life projectile motion cannot be perfectly modeled using equations. This is because there are many factors that can affect the motion, such as air resistance, wind, and variations in the initial conditions. However, equations and mathematical models can still provide a close approximation and help us understand and predict the motion of objects in real life.

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