470,127 views
9 votes
9 votes
A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a horizontal distance of 11 m, it has dropped 59 cm from its original height when it left the racquet. What was the initial speed, in m/s, of the tennis ball? (Neglect air resistance) m/s

User Marlyyy
by
2.4k points

1 Answer

5 votes
5 votes

\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=(gt^2)/(2) \\ \text{Solving t} \\ gt^2=2y \\ t^2=(2y)/(g) \\ t=\sqrt{(2y)/(g)} \\ t=\sqrt{(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=(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}

User Sz Ashik
by
2.8k points