Question

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 above the floor in an attempt to get the ball?

Short answer

$4.95\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step-by-step solution

Given Data

• The acceleration of the player,$a=-g=-9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.
• Final velocity, V= 0.
• Distance, $d=1.25m$

Velocity of the player

If an object starts with velocity U and after a distance d velocity changes to V, the achieved acceleration is a. The equation of motion can be written as:

${\mathbf{V}}^{\mathbf{2}}\mathbf{-}{\mathbf{U}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{gd}$

Substituting values in the above equation, we get:

$\begin{array}{rcl}{\left(0\right)}^2-{\left(U\right)}^2& =& 2\left(-9.81\right)\times \left(1.25\right)\\ U^2& =& 24.525\\ U& =& \sqrt{24.525}\\ U& =& 4.95\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

Hence the velocity at which the player should jump is $4.95\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.