A stone is thrown vertically upward. On its way up it passes point A with speed of v, and point B, 3 m higher than A, with speed v/2. The maximum height reached by stone above point B is

1 m

The correct answer is Option (1) → 1 m

$(\frac{v}{2})^2=v^2-2g×3$

$∴v=\sqrt{8g}$

If h is the further height, then

$0=(\frac{v}{2})^2-2gh$

$∴h=\frac{v^2}{8g}=\frac{8g}{8g}=1 m$