Answer
$v_0 = 38~ft/s$
Work Step by Step
On the graph, we can see that the ball moves a horizontal distance of $40~ft$ in a time of $1.25~s$. We can find $v_x$:
$v_x = \frac{40~ft}{1.25~s} = 32~ft/s$
On the graph, we can see that the shot reaches maximum height in a time of $0.625~s$
We can find $v_{0y}$:
$0.625~s = \frac{v_{0y}}{g}$
$v_{0y} = (g)~(0.625~s)$
$v_{0y} = (32~ft/s^2)~(0.625~s)$
$v_{0y} = 20~ft/s$
We can find the initial speed of the ball:
$v_0 = \sqrt{(32~ft/s)^2+(20~ft/s)^2} = 38~ft/s$
