# College Physics 2.43 – Basketball player’s velocity in a starting tip-off

## Solution:

The given values are

$v=0\:m/s$

$\Delta y=1.25\:m$

$a=-9.80\:m/s^2$

Therefore, the initial velocity of the player should be

$v^2=v_0^2+2a\Delta \:y$

$v_0^2=v^2-2a\Delta \:y$

$v_0=\sqrt{v^2-2a\Delta y}$

$v_0=\sqrt{\left(0\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.25\:m\right)}$

$v_0=4.95\:m/s$