Answer
The free-fall acceleration is $~~8.0~m/s^2$
Work Step by Step
From the graph, we can see that the ball reaches a maximum height of $25.0~m$ in a time of $2.5~s$
We can write an expression for the initial velocity $v_0$:
$v = v_0+at$
$0 = v_0+2.5~a$
$v_0 = -2.5~a$
We can find the free-fall acceleration $a$:
$y = v_0~t+\frac{1}{2}at^2$
$25.0 = (-2.5~a)~(2.5)+\frac{1}{2}a(2.5)^2$
$25.0 = -6.25~a+3.125~a$
$a = -\frac{25.0}{3.125}$
$a = -8.0~m/s^2$
The free-fall acceleration is $~~8.0~m/s^2$.
