# 2D Kinematics - Projectile Motion

## Homework Statement

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 55.9° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 29.5 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

## Homework Equations

Eight Kinematic Equations

## The Attempt at a Solution

Voy = VoSin55.9 = 75 sin 55.9 = 62.1045251
v^2 = Vy^2 + Vx^2
V^2 = 62.1^2 + 29.5^2

Definitely not right, I just can't seem to figure it out

Related Introductory Physics Homework Help News on Phys.org
So you found the vertical component of the initial velocity.
In the last line you try to add velocities and distances which doesn't make sense. Care to try another attempt?
Remember to break up two dimensional motion into two problems of one dimension each. Work in the vertical direction separately from the horizontal direction.

hage567
Homework Helper
You haven't taken gravity into account. How is that going to affect the trajectory of the rocket?
Also, look at the information in the x direction. You know there is no acceleration in that direction. Since you know the distance to the wall and Vx (although I think you made an error it isn't 29.5 m/s check your math) what can you find out using that?

29.5 was the given in the question.

Is the answer looking for the magnitude of the problem when it is right above the wall?

hage567
Homework Helper
29.5 was the given in the question.

Is the answer looking for the magnitude of the problem when it is right above the wall?
Yeah, but the 29.5 is distance in meters, not velocity in meters/second. So it is not Vx.

hi RKNY! I hope this helps,

X (wall) = 29.5m
Vox = (Vo)(cos)(theta)

to obtain the time while the rocket was exactly above the wall:
t = X / Vox

Y (abovewall) = Yo + Voyt - (4.9)(t)(t)

You do not include the height of the wall since you were asked only for the distance that was cleared by the rocket from the top of the wall. Cheers! :)