Physics, 29.11.2020 14:40 xxaurorabluexx

During a tennis match, a player serves the ball at 23.6 m/s, with the center of the ball leaving the racquet horizontally 2.37 m above the court surface. The net is 12 m away and 0.90 m 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 before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the net,
(c) does the ball clear it and
(d) what now is the distance between the center of the ball and the top of the net?

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Answer from: lizarroyo

Yes it clears

$b= 0.19 \ m$

No it does not clear

$z= 0.86 \ m$

Explanation:

From the question we are told that

The speed at which the player serves the ball is $v = 23.6 \ m/s$

The height of the ball above the ground is $h = 2.3 7 \ m$

The distance of the net is $d = 12 \ m$

The height of the net is $H = 0.9 \ m$

Generally the time taken for the ball to reach the net is mathematically represented as

$t = \frac{d}{v}$

=> $t = \frac{12}{23.6}$

=> $t = 0.508 \ s$

Generally the change in height of the ball after t is mathematically represented as

$\Delta h = ut + \frac{1}{2} gt^2$

Here u is the initial velocity which is zero given that the ball was at rest initially

$\Delta h = 0* t + \frac{1}{2} * 9.8 * 0.50 8^2$

=> $\Delta h =1.28 \ m$

Generally the new height of the ball is mathematically evaluated as

$s= h-\Delta h$

=> s = 2.37 - 1.28

=> $s = 1.09 \ m$

From the value obtained we see that s H hence the ball clears the net

Generally the distance between the center of the ball and the top of the net is mathematically represented as

b = s - H

=> b = 1.09 - 0.90

=> $b= 0.19 \ m$

Given that the ball makes an angle of 5^o with the horizontal , the velocity along the x-axis is

v_x = v cos(5)

=> v_x = 23.6 cos(5)

=> $v_x = 23.5 \ m/s$

The velocity along the y-axis is

v_y = v sin(5)

=> v_y = 23.6 sin(5)

=> $v_y = 2.06 \ m/s$

Generally the time taken for the ball to reach the net is

$t = \frac{d}{v_x}$

=> $t = \frac{12}{23.5}$

=> $t =0.508 \ s$

Generally the change in height of the ball after t seconds is

$c = v_yt + \frac{1}{2}gt^2$

=> $c = 2.06 * 0.508 + \frac{1}{2}* 9.8 * 0.508 ^2$

=> c = 2.33

Generally the new height of the ball after time t seconds is

e = h - c

=> e = 2.37 - 2.33

=> $e = 0.04 \ m$

From the value obtained we see that e < H hence the ball does not clear the net

Generally the distance between the center of the ball and the top of the net is mathematically represented as

z = H-e

=> z = 0.90 - 0.04

=> $z= 0.86 \ m$

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