College Physics by Openstax Chapter 2 Problem 43

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*}

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