Question

From an elevated point P a stone is projected vertically upward. When it reaches a distance $y$ below the point of projection its velocity is double the velocity when it was at a height $y$ above P. The greatest height reached by it above P is:

• A. $\frac{2y}{3}$
• B. $\frac{5y}{3}$
• C. $\frac{y}{3}$
• D. $2y$

Answer 1

Answer: (B)

$\frac{5y}{3}$

At point A ($y$ distance above): $v^2=u^2-2gy$ ....................... (1)
At point B ($y$ distance below): $(2v{)}^2=u^2+2gy$ .......................... (2)
Dividing eqn. (1) with (2), we get:
$\Rightarrow \frac{1}{4}=\frac{u^2-2gy}{u^2+2gy}$
$\Rightarrow u^2+2gy=4u^2-8gy$
$\Rightarrow 3u^2=10gy\Rightarrow u=\sqrt{\frac{10gy}{3}}$
So, maximum height above P is:

$\frac{u^2}{2g}=\frac{10gy}{3\times 2g}=\frac{5}{3}y$