Question

A tennis player throws a ball vertically upwards at 4.75 m/s. How high does the ball raise above their hand (assuming the hand has not moved since releasing the ball)?

Assume a gravitational acceleration of g = 9.8 m/s2 and give your answer in meters to 2 decimal places.

Answer 1

The motion of the body is described by the equations

$v_y=v_0-g \cdot t$ (1)

$y=v_0 \cdot t-\frac{g \cdot t^2}{2}$ (2)

when

$y=h_{max} \implies v_y=0$

equations take the form

$0=v_0-g \cdot t$ (1)

$h_{max}=v_0 \cdot t-\frac{g \cdot t^2}{2}$ (2)

then from equation (1) we write

$t= \frac{v_0}{g}$

substitute in (2) we get

$h_{max}=v_0 \cdot {\frac{v_0}{g}}-\frac{g \cdot {(\frac{v_0}{g}})^2}{2}= \frac{v_0^2}{g}-\frac{v_0^2}{2 \cdot g}=\frac{v_0^2}{2 \cdot g}=\frac{4.75^2}{2 \cdot 9.8}[m]=1.151[m]$