A few years ago, Serena Williams dived to hit a tennis ball right after it bounced off the ground. The ball bounced on the ground 10.3 m from the net, and after Serena hit the ball it flew over the 0.950 m high net and bounced in her opponent's court about 0.970 s after she hit it. If there had been no gravity, the ball would have been 1.36 m higher than the net when it crossed over. How fast was the ball moving when it left Serena/'s racket? Ignore any effects from ball spin and air resistance.

Question

Zachariah Chan

Physics

30 October, 23:30

A few years ago, Serena Williams dived to hit a tennis ball right after it bounced off the ground. The ball bounced on the ground 10.3 m from the net, and after Serena hit the ball it flew over the 0.950 m high net and bounced in her opponent's court about 0.970 s after she hit it. If there had been no gravity, the ball would have been 1.36 m higher than the net when it crossed over. How fast was the ball moving when it left Serena/'s racket? Ignore any effects from ball spin and air resistance.

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Mcgrath · Answer 1

is possibly approximate, as maximum height attained may have been over the top of the net, i. e. small error in horizontal V component. Or maybe comes from the "about" 1.15 secs. in air.

Explanation:

2.3 + 0.95) = 3.25 metres.

Angle of hit = arctan (3.25/11.3) = 16.046 degrees above horizontal.

The ball was 1.15 secs. in the air, so reached max. height in (1.15/2) = 0.575 sec.

Height attained over court = 1/2 (t^2 * g) = 1.62 metres.

Initial vertical V component = sqrt. (2gh) = sqrt. (19.6 x1.62) = 5.6349m/sec.

Range = (4h/tan 16.046) = (1.62 x 4) / tan 16.046 = 22.53 metres.

Horizontal V component = (22.53/1.15s) = 19.59m/sec.

Speed of hit = sqrt. (19.59^2 + 5.6349^2) = 20.384m/sec.