Posted by shawni on Tuesday, September 11, 2012 at 9:28pm.
V^2/2 increases by g*H m2/s^2, the potential energy loss.
The time in the air is the sum of the times spent going up and down.
The time going up is t1 = 7.79 m/s/g
The time spent coming down is given by
Vyo*t2 -(g/2)*t2^2 = -14.3 meters
Solve for t2
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