## Statement

A tennis ball located 2 m height is hit by a player with his racket. The ball shoots out horizontally with a velocity of 30 m/s. Answer the following questions:

a) How long does the ball takes to hit the ground?

b) What is the angle of the velocity vector with the x-axis at the time the ball reaches the ground?

c) If before being hit, the ball is 5 meters away from the net, how high does the ball pass above the net?

## Solution

Question a)

The ball will reach the ground when the y-coordinate of its position is 0 (y=0). According to the equations of horizontal launch:

$$y=H-\frac{1}{2}\xb7g\xb7{t}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}0=2m-0.5\xb79.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${s}^{2}$}\right.\xb7{t}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)t=0.63s}$$

Question b)

To calculate the angle that the velocity vector forms with the x-axis, we will use the following expression:

$$\mathrm{tan}\left(\alpha \right)=\frac{{v}_{y}}{{v}_{x}}$$

To solve it, we will calculate v_{x} and v_{y}:

$${v}_{x}={v}_{0}=\overline{)30\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\phantom{\rule{0ex}{0ex}}{v}_{y}=-g\xb7t\Rightarrow {v}_{y}=-9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${s}^{2}$}\right.\xb70.63s=-\overline{)6.17\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}$$

Once we have the velocity data, we can get the angle:

$$\mathrm{tan}\left(\alpha \right)=\frac{-6.17\overline{){\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}}{30\overline{){\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}}\Rightarrow \phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\alpha \right)=-0.205\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)\alpha =-11.62\xb0}$$

Question c)

To calculate the height at which the ball will pass above the net, first, we must know when it crosses above the net. To do so, knowing that the ball is 5 m away from the net and moves with a u.r.m. at 30 m/s:

$$x={v}_{0}\xb7t\Rightarrow 5\overline{)m}=30\raisebox{1ex}{$\overline{)m}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\xb7t\Rightarrow \overline{)t=0.17s}$$

At that time the ball will be just over the net, so we simply calculate its position and we would have solved the problem:

$$y=H-\frac{1}{2}\xb7g\xb7{t}^{2}\Rightarrow 2-0.5\xb79.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${s}^{2}$}\right.\xb7(0.17s{)}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)y=1.86m}$$