Answer
The magnitude of the free-fall acceleration is $~~8.0~m/s^2$
Work Step by Step
On the graph, we can see that the shot reaches maximum height in a time of $1.25~s$
We can find an expression for $g$:
$1.25~s = \frac{v_{0y}}{g}$
$g = \frac{v_{0y}}{1.25~s}$
The shot moves up a vertical distance of $4.0~m$ in a time of $0.50~s$. We can find $v_{0y}$:
$y = y_0+v_{0y}~t-\frac{1}{2}gt^2$
$y-y_0 = v_{0y}~t-\frac{1}{2}(\frac{v_{0y}}{1.25~s})(t)^2$
$4.0~m = v_{0y}~(t-\frac{t^2}{2.5~s})$
$v_{0y} = \frac{4.0~m}{t-\frac{t^2}{2.5~s}}$
$v_{0y} = \frac{4.0~m}{0.50~s-\frac{(0.50~s)^2}{2.5~s}}$
$v_{0y} = 10~m/s$
We can find $g$:
$g = \frac{v_{0y}}{1.25~s}$
$g = \frac{10~m/s}{1.25~s}$
$g = 8.0~m/s^2$
The magnitude of the free-fall acceleration is $~~8.0~m/s^2$
