Question

A ball is thrown vertically up from the point $A$ (See figure). A person, standing at a height $H$ on the roof a building, tries to catch it. He misses the catch, the ball overshoots and simultaneously the person starts a stop-watch. The ball reaches its highest point and he manages to catch it upon its return. By this time, a time interval $T$ has elapsed as recorded by the stop watch. If $g$ is the acceleration due to gravity at this place, the speed with which the ball was thrown from point $A$ will be

• A. $\sqrt{gH+gT}$
• B. $\frac{\left(\sqrt{g^2{T}^2+4gH}\right)}{2}$
• C. $\frac{\left(\sqrt{g^2{T}^2+8gH}\right)}{2}$
• D. $\left(\sqrt{g^2{T}^2+2gH}\right)$

Answer 1

The correct option is C $\frac{\left(\sqrt{g^2{T}^2+8gH}\right)}{2}$

Let the velocity of ball near roof top be 'v'

By kinematics equations,

$v=u+at$

$0=v^{\prime }-gT/2$

$v^{\prime }=\frac{gT}{2}$

By applying $v^2=u^2-2gH$

$\frac{g^2{T}^2}{4}=u^2-2gH$

$u=\frac{\sqrt{g^2{T}^2+8Hg}}{2}$