Answer
The break-free point is at a height of $~~58.8~m$
Work Step by Step
From the graph, we can see that the break-free point occurs at $t = 2.0~s$ when $v = 19.6~m/s$
At $t = 8.0~s$, the package reaches the ground. Thus, after breaking free, the package takes 6.0 seconds to reach the ground.
We can find the height $y_0$ of the break-free point:
$y-y_0 = v_0~t+\frac{1}{2}at^2$
$0-y_0 = (19.6~m/s)(6.0~s)+\frac{1}{2}(-9.8~m/s^2)(6.0~s)^2$
$-y_0 = -58.8~m$
$y_0 = 58.8~m$
The break-free point is at a height of $~~58.8~m$.
