91影视

In this problem, let us assume the origin to be the location on the ground underneath the ball when it is served.In order to find the time required by the ball to reach the net, the equation of motion will be utilized.

Given data:

The height of the net is $h=0.9\mathrm{m}$.

The distance from the serve is $d=15\mathrm{m}$.

The height from which the ball is launched is $h_b=2.50\mathrm{m}$.

The distance at which the ball lands is $d_b=7\mathrm{m}$.

The relation from the equation of motion is given by:

$h=h_b+ut+\frac{1}{2}g{t}^2$

Here, u is the speed of the ball in the vertical direction whose value is zero, t is the time, and g is the gravitational acceleration.

On plugging the values in the above relation, you get:

$\begin{array}{c}0.9鈥�\mathrm{m}=2.50鈥�\mathrm{m}+\left(0\right)t+\frac{1}{2}\left(-9.81\mathrm{m}/{\mathrm{s}}^2\right)t^2\\ t=0.57\mathrm{s}\end{array}$

The formula to calculate the speed can be written as:

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

On plugging the values in the above relation, you get:

$\begin{array}{c}v=\left(\frac{15\mathrm{m}}{0.57\mathrm{s}}\right)\\ v=26.3\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v=26.3\mathrm{m}/\mathrm{s}$is the minimum speed of the ball.

The formula to calculate the speed can be written as:

$y=h_b+ut\text{'}+\frac{1}{2}gt{\text{'}}^2$

Here, y is the final location of the ball whose value is zero, and t鈥� is the time of flight of the ball.

On plugging the values in the above relation, you get:

$\begin{array}{c}0=2.50\mathrm{m}+\left(0\right)t\text{'}+\frac{1}{2}\left(-9.81\mathrm{m}/{\mathrm{s}}^2\right)t{\text{'}}^2\\ t\text{'}=0.71\mathrm{s}\end{array}$

The formula to calculate the horizontal position can be written as:

$x=vt\text{'}$

On plugging the values in the above relation, you get:

$\begin{array}{l}x=\left(26.3\mathrm{m}/\mathrm{s}\right)\left(0.71\mathrm{s}\right)\\ x=18.6\mathrm{m}\end{array}$

Thus, $x=18.6\mathrm{m}$is the horizontal position. Also, the ball will land in the good region because its position is between the 15 m and 22 m region.