Afew 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.

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.

the answer is 3.1 his answer is wrong mines is right