[tex]\begin{gathered} x=11m \\ y=59\operatorname{cm}=0.59m \\ g=9.81m/s^2 \\ v_o=?\text{ in m/s} \\ \text{When the ball }leaves\text{ the ra}cquet,\text{ it moves }horizontally,\text{ hence} \\ v_y=0\text{ } \\ \text{then} \\ y=\frac{gt^2}{2} \\ \text{Solving t} \\ gt^2=2y \\ t^2=\frac{2y}{g} \\ t=\sqrt{\frac{2y}{g}} \\ t=\sqrt{\frac{2(0.59m)}{9.81m/s^2}} \\ t=0.35s \\ \text{After 0.35s, the ball has traveled 11m, horizontaly and dropped 59cm} \\ \text{hence} \\ \text{the follow}ing\text{ equation d}escribes\text{ the moving of the ball horizontaly} \\ x_{}=v_ot \\ \text{Solving v}_o \\ v_o=\frac{x}{t} \\ v_o=\frac{11m}{0.35\text{ s}} \\ v_o=31.43\text{ m/s} \\ \text{The initial sp}eed\text{ of the ball is }31.43\text{ m/s} \end{gathered}[/tex]
