Question

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

• A. $\frac{5y}{4}$
• B. $\frac{5y}{3}$
• C. $\frac{y}{3}$
• D. $2y$

Answer 1

The correct option is B $\frac{5y}{3}$

Let the initial velocity = u

At point 'P'

$AP=h\Rightarrow V_p^2-u^2=-2gh.......\left(i\right)$

At point 'Q'

$V_Q^2-u^2=2(-g)(-h)\Rightarrow V_Q^2-u^2=2gh$

As $V_Q=2V_P$

$4V_P^2-u^2=2gh........\left(ii\right)$

$4[u^2-2gh]-u^2=2gh$

$4u^2-8gh-u^2=2gh$

$3u^2=10gh$

$u^2=\frac{10}{3}gh.......\left(iii\right)$

But $0^2-u^2=-2gh........\left(iv\right)$

$2gH=\frac{10}{3}gh$

$2H=\frac{10}{3}h$

$H=\frac{5}{3}h$

According to the question height is 'y' so

$H=\frac{5}{3}y$