A Cutnell and Johnson Physics question illustrates a rocket launch:(adsbygoogle = window.adsbygoogle || []).push({});

A rocket is fired at a speed of 97.0 m/s from ground level, at an angle of 53.0 ° above the horizontal. The rocket is fired toward an 27.0-m high wall, which is located 24.0 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?

My Calculations:

Vo = 97.0 m/s

Theta = 53 degrees

x = 24 m

y = 27 m

Let H illustrate the height of the rocket at point x on the axis.

t = ?

a = -9.80 m/s ^2

Vo sin 53 = Voy = 77.46

Vo cos 53 = Vox = 58.37

x = Vox(t)

t = x/Vox = 24/58.37 = .411 s

H = Voy(t) + .5 a(t^2)

Plug and Chug

H = 77.46(.411) + .5(-9.80)(.411^2) = 31 m

Distance Rocket clears wall = 31m-27m = 4.00 m <-----Incorrect.

Is there a problem with my calculations that I am not understanding?

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# Homework Help: Projectile Motion, A study of height and time

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