Answer
The maximum height is $~~9.3~ft$
Work Step by Step
On the graph, we can see that the shot reaches maximum height in a time of $~~0.625~s$
In part(a), we found that $~~v_{0y} = 20~ft/s$
We can find the maximum height:
$y = y_0+v_{0y}~t-\frac{1}{2}gt^2$
$y = (3.0~ft)+(20~ft/s)(0.625~s)-\frac{1}{2}(32~ft/s^2)(0.625~s)^2$
$y = 9.3~ft$
The maximum height is $~~9.3~ft$
