Tennis and the horizontal launch

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:

$$\begin{gathered} y=H-\frac{1}{2}·g·t^2\Rightarrow \\ 0=2 m-0.5·9.8 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.·t^2\Rightarrow \\ \boxed{t=0.63 s} \end{gathered}$$

Question b)

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

$$\tan \left(\alpha \right)=\frac{v_y}{v_x}$$

To solve it, we will calculate v_x and v_y:

$$\begin{gathered} v_x=v_0=\underline{30 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.} \\ {v}_y=-g·t \Rightarrow v_y=-9.8\raisebox{1ex}{$ m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.·0.63s=-\underline{6.17 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.} \end{gathered}$$

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

$$\begin{gathered} \tan \left(\alpha \right)=\frac{-6.17 \cancel{{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}}{30 \cancel{{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}}\Rightarrow \\ \tan \left(\alpha \right)=-0.205 \Rightarrow \\ \boxed{\alpha =-11.62°} \end{gathered}$$

 

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:

$$x=v_0·t \Rightarrow 5 \cancel{m}=30 \raisebox{1ex}{$\cancel{m}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.·t \Rightarrow \underline{t=0.17 s}$$

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

$$\begin{gathered} y=H-\frac{1}{2}·g·t^2\Rightarrow 2-0.5·9.8 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.·(0.17 s{)}^2\Rightarrow \\ \boxed{y=1.86 m} \end{gathered}$$