Posted by Bobby on Tuesday, October 24, 2006 at 6:45pm.
During a tennis match a player serves the ball at 23.6 m/s with the center of the ball leaving the raquet horizontally 2.37m abovet eht court surface. The net is 12m away and .9m high. When the ball reaches the net a) does the ball clear it and b) what is the distance between the center of the ball and the top of the net?
Suppose that instead the ball is served as befor but now it leaves the racquet at 5 degrees below the horizontal. When the ball reaches the net, c) does the ball clear it and d) what is the distance between the center of the ball and the top of the net?
This problem was wordy and it confused me.
First part: How long does it take the ball to get to the net (time=distane/velocity).
Using that time, how far does it fall vertically?
distance=1/2 g t^2
On the second, you have an initial vertical velocity,that will figure into the vertical fall.

PHYSICS  Anonymous, Monday, October 8, 2012 at 6:46pm
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