A rocket is fired at a speed of 75.5 m/s from ground level, at an angle of 62.0° above the horizontal. The rocket is fired toward an 12.0 m high wall, which is located 28.7 m away. By how much does the rocket clear the top of the wall?

Question

anonymous · Answer

first determine the x and y component of initial velocity

anonymous · Answer

Right, I tried that and then used the x components to determine the amount of time it would take to cover 28.7 m and used that time to find how high the rocket would be at the point it is over the wall. However, my answer was incorrect and I am unsure why.

anonymous · Answer

What did you get as your answer?

anonymous · Answer

206.5m once I subtracted the 12m of the wall

anonymous · Answer

what did you get as the time?

anonymous · Answer

t=8.097s

anonymous · Answer

your time is off by a factor of 10

anonymous · Answer

the horizontal distance between teh wall and the rocket is 28.7 m

the horizontal velocity, i calculated to be 35.34 m/s

anonymous · Answer

d=v*t
t= d/v

anonymous · Answer

$$v_{xo} = 75.5 \cos(62.0 ^\circ)$$

anonymous · Answer

I can't read my own handwriting apparently, I just missed a decimal point when I typed it into my calculator. Thanks for helping me catch that mistake

anonymous · Answer

no problem