2D Kinematics - Projectile Motion

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Homework Help Overview

The problem involves analyzing the projectile motion of a rocket fired at a specific speed and angle, with the goal of determining how much it clears a wall of a given height located a certain distance away. The subject area is 2D kinematics, specifically focusing on projectile motion and the effects of gravity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the vertical and horizontal components of the rocket's motion, with some questioning the original poster's calculations and assumptions. There are suggestions to separate the motion into distinct vertical and horizontal analyses. Additionally, gravity's role in the trajectory is highlighted, prompting further inquiry into the relationship between distance and velocity.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem. There is an emphasis on clarifying the distinction between distance and velocity, and some participants are exploring the implications of the given data on the rocket's trajectory.

Contextual Notes

Participants note that the original poster may have misunderstood the relationship between the given distance to the wall and the calculations involving velocity. There is also mention of the need to account for gravity in the trajectory analysis.

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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