Posted by **hy** on Sunday, October 24, 2010 at 2:10pm.

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 64.7¡ã above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 25.0 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?

- physics -
**bobpursley**, Sunday, October 24, 2010 at 2:18pm
Horizontal

distance=horzvelocity*time

time= distance/velocity= 25/(75cos64.7)

figure time, then put it here.

Vertical:

h=75*sin64.7t-4.9t^2

how much higher is h than 11 m.

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**hy**, Sunday, October 24, 2010 at 2:33pm
thanks

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**liz**, Sunday, January 16, 2011 at 3:50pm
where does the 4.9 come from?

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**tom**, Wednesday, September 21, 2011 at 2:18am
h=49.62m

49.62m-11m for the wall

clear of the top = 38.62

4.9 its from multiplying 1/2(-9.8m/s^2)

1/2(ay)

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