Question

In a tennis match, a tennis player serves the ball from 2.37 m above the ground at an angle of 5.0 degrees below horizontal. The ball is served from the baseline 12.0 m from the net. (a) If the ball clears the 0.90 m high net by 8.0 cm, how fast was the ball served? (b) Given that the opponent’s baseline is also 12.0 m beyond the net, does the ball land “in” or “out”, and by how much? (Assume no air-resistance, of course.)

Answer #1

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with the center of the ball leaving the racquet horizontally 2.37 m
above the court surface. The net is 12.0 m away and 0.900 m high.
When the ball reaches the net, (a) what is the
distance between the center of the ball and the top of the net?
(b) Suppose that, instead, the ball is served as
before but now it leaves the racquet at 5.00° below...

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(b) Suppose that, instead, the ball is served as
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Thank you so much in advance!

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