# (solved) Why is my answer wrong? Projectile Motion

## 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...

## Answers and Replies

Chestermiller
Mentor
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 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 ?...

#### Attachments

• tati1.PNG
27 KB · Views: 664
Last edited:
Chestermiller
Mentor
Looks OK.