Question

A tennis player hits a ball 2.0 m above the ground. The ball

leaves his racquet with a speed of 20.0 m/s at an angle 5.0° above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Does the ball clear the net? If so, by how much? If not, by how much does it miss?

Short answer

Yes, the ball clear the net by$1.0\text{m}$.

Step-by-step solution

Step1. Given information

The initial vertical position of the ball is $2.0\text{m}$, the initial speed of the ball is $20.0\text{m/s}$, the angle of projection is $5.0°$, the horizontal distance travelled by the ball is$7.0\text{m}$and the height of the net is$1.0\text{m}$.

Step 2. Calculation of the time taken by the ball

The formula to calculate the horizontal distance travelled by the ball is given by

$x=v_0\cos \theta t...................\left(1\right)$

Here, $x$is the horizontal distance travelled, $v_0$is the initial speed of the ball, $\theta$is the projection angle and $t$is the time required.

Substitute $20.0\text{m/s}$for $v_0$, localid="1647486771912" $5.0°$for $\theta$and $7.0\text{m}$for $x$into equation (1) and solve to calculate the time required.

$\begin{array}{rcl}7.0\text{m}& =& 20.0\text{m/s}\times \cos 5.0°\times t\\ t& =& \frac{7.0\text{m}}{20.0\text{m/s}\times \cos 5.0°}\\ & \approx & 0.35\text{s}\end{array}$

Step 3. Calculation of the final vertical height

The formula to calculate the final vertical height of the ball is given by

$y=y_0+v_0\sin \theta t-\frac{1}{2}g{t}^2....................\left(2\right)$

Here, $y,y_0,g$are the final vertical height, initial vertical height and acceleration due to gravity respectively.

Substitute $2.0\text{m}$for $y_0$, $20.0\text{m/s}$for $v_0$, $5.0°$for $\theta$, $0.35\text{s}$for $t$and $9.80{\text{m/s}}^2$for $g$into equation (2) to calculate the final vertical height of the ball.

$\begin{array}{rcl}y& =& 2.0\text{m}+20.0\text{m/s}\times \sin 5.0°\times 0.35\text{s}-\frac{1}{2}\times 9.80{\text{m/s}}^2\times {\left(0.35\text{s}\right)}^2\\ & \approx & 2.0\text{m}\end{array}$

The formula to calculate the distance$\left(∆y\right)$ between the net and the final vertical position of the ball is given by

$∆y=y-l....................\left(3\right)$

Here, $l$is the height of the net.

Substitute $2.0\text{m}$for $y$and $1.0\text{m}$for $l$into equation (3) to calculate the required difference in height.

$\begin{array}{rcl}∆y& =& 2.0\text{m}-1.0\text{m}\\ & =& 1.0\text{m}\end{array}$