2D Kinematics - Projectile Motion

AI Thread Summary
A rocket is launched at 75.0 m/s at a 55.9° angle, aiming to clear an 11.0 m high wall located 29.5 m away. The vertical component of the initial velocity is calculated to be approximately 62.1 m/s. Participants emphasize the need to separate the motion into horizontal and vertical components, considering gravity's effect on the trajectory. The horizontal distance to the wall is used to determine the time of flight, while the vertical motion equation helps find the rocket's height at that point. The final goal is to determine how much the rocket clears the wall, focusing solely on the distance above the wall's height.
RKNY
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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
 
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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.
 
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?
 
RKNY said:
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! :)
 
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