Projectile motion rocket problem

AI Thread Summary
A rocket is launched at 75.0 m/s at a 60-degree angle, aiming to clear an 11.0 m high wall located 27.0 m away. The initial vertical velocity is calculated as 65.0 m/s, and the horizontal velocity is 37.5 m/s. The time to reach the wall is determined to be 0.720 seconds, resulting in a vertical position of 49.3 m at that time. The discrepancy in the expected answer of 33.2 m suggests an error in accounting for vertical acceleration, which should be -9.8 m/s². Correcting this will yield the accurate height above the wall.
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Homework Statement


A rocket is fired at at speed fo 75.0 m/s from ground level, at an angle of 60.0 deg. above the horizontal. The rocket is fired toward an 11.0 m high wall, shich is located 27.0 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?


Homework Equations





The Attempt at a Solution


I got the initial vy by 75.0(sin(60))=65.0 m/s, and vx 75.0(cos(60))=37.5 m/s
27.0m=(37.5m/s)t + (1/2*0*t)
t=0.720s
y=65.0 m/s(0.720s) + 1/2(9.80 m/s^2)(0.720^2)
y=49.3m
49.3-11.0=38.3m above the wall
The answer is supposed to be 33.2m, I'm not sure where I went wrong.
 
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vertical acceleration = -9.8m/s^2 not 9.8m/s^2
 
thanks
 
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