Answer
$v\approx10m/s$ (other approximations are also possible)
Work Step by Step
Assuming that the apple has fallen 4 m before hitting Newton's head:
Let the positive direction be downward:
$v^2=v_{0}^{2}+2g\Delta x$. But, $v_{0}=0$
$v=\sqrt {2g\Delta x}=\sqrt {2(9.81~m/s^2)(4~m)}$
$v\approx10m/s$
