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


A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?


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