Question

A body $A$ is thrown up vertically from the ground with velocity $v_o$ and another body $B$ is simultaneously dropped from a height $H$. They meet at a height $\frac{H}{2}$ if $v_o$ is equal to

• A. $\sqrt{2gH}$
• B. $\sqrt{gH}$
• C. $\frac{1}{2}gH$
• D. $\sqrt{\frac{2g}{H}}$

Answer 1

The correct option is B $\sqrt{gH}$

time taken by first body to reach $\frac{H}{2}$ is t

& time taken by second body to fall a

height $\frac{H}{2}$ is $t_1$ then both time should

be equal so that both reach the

height $\frac{H}{2}$ simultaneously

For second body $h=\frac{1}{2}g{t}^2$

gives $\frac{H}{2}=\frac{1}{2}g{t}_1^2\Rightarrow t_1^2=\frac{H}{g}$

For first body $h=ut-\frac{1}{2}g{t}^2$

gives $\frac{H}{2}=v_{0}t-\frac{1}{2}g{t}^2$

put $t=t_1$ we get $\frac{H}{2}=V_0\sqrt{\frac{H}{g}}-\frac{1}{2}.g.\frac{H}{2}$

$\Rightarrow \frac{H}{2}=v_0\sqrt{\frac{H}{9}}-\frac{H}{2}\Rightarrow H=v_0\sqrt{\frac{H}{9}}$

$\Rightarrow v_0=\sqrt{gH}$