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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.

Answer :

Answer:

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 speed of the ball moving when it left Serena Williams racket is 21.23 m/s.

What is the speed of a body?

The speed of a body is the rate at which it covers the total distance is in the time taken.

A few years ago, Serena Williams dived to hit a tennis ball right after it bounced off the ground.

As, the initial vertical position is zero. Thus, the vertical component of velocity can be given as,

[tex]V_{yo}\times t=\dfrac{1}{2}gt^2\\V_{oy}\times (0.970)=\dfrac{1}{2}(9.81)(0.970)^2\\V_{yo}=4.758\rm m/s[/tex]

Here, 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.

Form, the trajectory of the ball, it has the straight line with the slope of,

[tex]\dfrac{\Delta y}{\Delta x}=\dfrac{0.95 m (y)+1.36m (y)}{10.3 m (x)}\\\dfrac{\Delta y}{\Delta x}=0.23\dfrac{m(y)}{m (x)}[/tex]

Hence, the ball goes 0.23 meters above the ground for each meter it travels in the right direction.

If the speed of the ball is 4.758 m/s then, the horizontal component of velocity can be given as,

[tex]V_x=\dfrac{4.758}{0.23}\\V_x=20.69\rm m/s[/tex]

The magnitude of the initial speed is,

[tex]V_o=\sqrt{20.69^2+4.76^2}\\V_o=21.23\rm m/s[/tex]

Thus, the speed of the ball moving when it left Serena Williams racket is 21.23 m/s.

Learn more about the speed here:

https://brainly.com/question/359790