Question

From an elevated point A , a ball is projected up. The speed of the stone at a depth 30 m below point A is double of its speed when it was at a height 30 m above A. The greatest height attained by the stone above A is :

• A. 50 m
• B. 60 m
• C. 100 m
• D. 35 m

Answer 1

The correct option is A 50 m

Let the initial velocity$=u$

At point ${}^{\prime }{P}^{\prime }$

$AP=h\Rightarrow V_P^2-u^2=-2gh\to 1$

$V_P^2=u^2-2gh$

At point Q

$V_Q^2-u^2=-2(-g)(-h)$

$V_P^2-u^2=+2gh$

As $V_Q=2V_P$

$4V_P^2-u^2=2gh\to 2$

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

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

$3u^2=10gh$

$u^2=\frac{10}{3}gh\to 3$

But $0^2-u^2=-2gH\to 4$

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

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

$=\frac{5}{3}\times 30$ [put $h=30$]

$H=50m$

The greatest height attained by the stone above A is $50m$