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:

It is our assumption that the player attempts to get the ball at the top where the velocity is zero.

The given are the following: v_{fy}=0 \ \text{m/s}; \Delta y=1.25 \ \text{m}; and a=-9.80 \ \text{m/s}^2.

We are required to solve for the initial velocity v_{0y} of the player. We are going to use the formula

\left(v_{fy}\right)^2=\left(v_{oy}\right)^2+2a\Delta y

Solving for v_{oy} in terms of the other variables:

v_{oy}=\sqrt{\left(v_{fy}\right)^2-2a\Delta y}

Substituting the given values:

\begin{align*}
v_{oy} & =\sqrt{\left(v_{fy}\right)^2-2a\Delta y} \\
v_{oy} & = \sqrt{\left(0\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.25\:\text{m}\right)} \\
v_{oy} & =4.95 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}