Question

A basketball player is practicing their standing vertical job. They jump 0.65 m off the ground because of their 0.15 m crouch (their tuck to generate their jump), which generated their force. If the player has a mass of 58 kg, find their jumping force.

Answer 1

The final velocity of the player can be expressed as,

$v^2=u^2-2gh$

At the maximum height the final velocity of player is zero. Plug in the known values,

$\begin{gathered} (0m/s)^2=u^2-2(9.8m/s^2)(0.65\text{ m)} \\ u^2=12.74m^2s^{-2} \\ u=3.57\text{ m/s} \end{gathered}$

Now, when the initial velocity becomes, the final velocity of player as it moves. The final velocity can be expressed as,

$v^2=u^2+2ah_0$

Substitute the known values,

$\begin{gathered} (3.57m/s)^2=(0m/s)^2+2a(0.15\text{ m)} \\ a=\frac{(3.57m/s)^2}{2(0.15\text{ m)}} \\ =42.5m/s^2 \end{gathered}$

The jumping force of the player is,

$F=ma$

Substuting values,

$\begin{gathered} F=(58kg)(42.5m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =2465\text{ N} \end{gathered}$

Thus, the jumping force acting on the player is 2465 N.