Question

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?

Answer 1

Answer:

a

Yes it clears

b

 $b= 0.19 \ m$

c

 No it does not clear

d

$z= 0.86 \ m$

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

So

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

=>  $\Delta h =1.28 \ m$

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

      $z = H-e$

=>   $z = 0.90 - 0.04$

=>   $z= 0.86 \ m$

   

Answer 2

(a) Yes, the ball clears the net.

(b) The distance between the center of the ball and the top of the net is 0.203 m.

(c) No, the ball does not clear the net.

(d) Now, the distance between the center of the ball and the top of the net is -0.85 m.

What is a Projectile motion?

When any object or body is launched with some initial velocity and making some angle with the horizontal, the body travels in a parabolic path. It is known as the projectile motion.

Given,

The horizontal distance traveled by the ball is 12 m.

The height of the top of the net is 0.90 m.

The height of the horizontal launch of the ball is 2.37 m.

The time for the horizontal motion of the projectile that is the ball is,

$\begin{aligned} {{v}_{0x}}&={{S}_{x}}t \\ t&=\dfrac{{{S}_{x}}}{{{v}_{0x}}} \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}\cos 0{}^\circ } \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}} \end{aligned}$

The equation for the vertical motion of the projectile can be solved by substituting the above result.

$\begin{aligned} y&={{y}_{0}}+{{v}_{0y}}t-\frac{1}{2}g{{t}^{2}} \\ y&={{y}_{0}}+\left( {{v}_{0}}\sin 0{}^\circ \right)\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)-\frac{1}{2}g{{\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)}^{2}} \\ \left( 0.90\text{ m}+h \right)&=\left( 2.37\text{ m} \right)+0-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( \frac{12\text{ m}}{23.6\text{ m/s}} \right)}^{2}} \\ h&=2.37\text{ m}-\text{1}\text{.267 m}-0.90\text{ m} \\ &=0.203\text{ m} \end{aligned}$

Now, consider the case when the ball is thrown with an angle $\bold{5^{\circ}}$ with the horizontal.

The horizontal component of the initial velocity is $\bold{v_0 \cos 5{}^\circ.}$

The vertical component of the initial velocity is $\bold{v_0 \sin 5{}^\circ.}$

For the horizontal distance traveled by the ball in this case, the time taken can be calculated as below,

$\begin{aligned} {t}'&=\frac{{{{{S}'}}_{x}}}{{{{{v}'}}_{0x}}} \\ &=\frac{12\text{ m}}{\left( 23.6\text{ m/s} \right)\cos 5{}^\circ } \\ &=0.51\text{ s} \end{aligned}$

Now, the vertical distance above the ground, y’, traveled by the projectile till reaching the net can be determined as,

$\begin{aligned} {y}'&={{{{y}'}}_{0}}+{{{{v}'}}_{0y}}{t}'-\frac{1}{2}g{{{{t}'}}^{2}} \\ &=0\text{ m}-\left( \left( 23.6\text{ m/s} \right)\sin 5{}^\circ \right)\left( 0.51\text{ s} \right)-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( 0.51\text{ s} \right)}^{2}} \\ &=-1.05\text{ m}-1.28\text{ m} \\ &=-2.32\text{ m} \end{aligned}$

The height above the top of the net can be determined by adding the above result with (2.37 m - 0.90 m) which is the height of the net’s top relative to the launch position.

$\begin{aligned} {h}'&=\left( 2.37\text{ m}-0.90\text{ m} \right)+{y}' \\ &=\left( 2.37\text{ m}-0.90\text{ m} \right)-2.32\text{ m} \\ &=-0.85\text{ m} \end{aligned}$

Thus, the distance between the center of the ball and the top of the net is -2,32 m.

When the ball leaves the racquet at 5.00° below the horizontal, the distance between the center of the ball and the top of the net is -0.85 m.

Learn more about projectile motion here:

https://brainly.com/question/11049671