From an elevated point $A$, a stone is projected vertically upwards. When the stone reaches a distance $h$ below $A$, its velocity is double of what was at a height above $A$. The greatest height attained by the stone is

Question

Dr. Abubacker · Accepted Answer

(d) Let $u$ be the velocity with which the stone is projected vertically upwards.
Given that, $v_{- h} = 2v_{h}$
$\left( v_{- h} ight)^{2} = 4v_{h}^{2}$
$	herefore\ u^{2} - 2g( - h) = 4(u^{2} - 2gh)$
$	herefore\ \ u^{2} = \frac{10gh}{3}$
Now, $h_{\max} = \frac{u^{2}}{2g} = \frac{5h}{3}$

Q. From an elevated point $A$, a stone is projected vertically upwards. When the stone reaches a distance $h$ below $A$, its velocity is double of what was at a height above $A$. The greatest height attained by the stone is

Explanation

Reviewed by Dr. Abubacker

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