91Ó°ÊÓ

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$1.0\text{m}$.

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

$x=v_0\cos \mathrm{θ}t...................\left(1\right)$

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

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

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

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

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

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 $\mathrm{θ}$, $0.35\text{s}$for $t$and $9.80{\text{m/s}}^2$for $g$into equation (2) to calculate the final vertical height of the ball.

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

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.

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