Q43PE - College Physics (Urone)

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?

Step-by-Step Solution

Answer

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

Step 1: Given Data

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

Step 2: 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$}\right.$ .