Answer
The maximum height reached above point B is $~~1.0~m$
Work Step by Step
The stone travels a distance of $y = 3.00~m$ between point A and point B.
The initial velocity at point A is $v$ and the final velocity at point B is $\frac{1}{2}v$
We can find $v$:
$v_f^2 = v_0^2+2ay$
$(\frac{1}{2}v)^2 = v^2+2ay$
$\frac{3}{4}v^2 = -2ay$
$v^2 = -\frac{8ay}{3}$
$v = \sqrt{-\frac{8ay}{3}}$
$v = \sqrt{-\frac{(8)(-9.8~m/s^2)(3.00~m)}{3}}$
$v = 8.854~m/s$
$\frac{1}{2}v = 4.43~m/s$
In the next part of the question, let $v_0 = 4.43~m/s$ and let $v_f = 0$
We can find the maximum height reached above point B:
$v_f^2 = v_0^2+2ay$
$y = \frac{v_f^2-v_0^2}{2a}$
$y = \frac{0-(4.43~m/s)^2}{(2)(-9.8~m/s^2)}$
$y = 1.0~m$
The maximum height reached above point B is $~~1.0~m$.
