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

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

Motion in a Straight Line

Solution:

$\left(\frac{v}{2}\right)^{2}=v^{2}-2 g \times 3$
$\therefore v=\sqrt{8 g}$
If $h$ is the further height, then
$0=\left(\frac{v}{2}\right)^{2}-2 g h$
$\therefore h=\frac{v^{2}}{8 g}=\frac{8 g}{8 g}=1\, m$