Explanation

(b) Let the two bodies $A$ and $B$ respectively meet at a time $t$, at a height $\frac{H}{2}$ from the ground
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Using $S = ut + \frac{1}{2}at^{2}$
For a body $A,\ u = V_{0},\ a = - g,\ S = \frac{H}{2}$
$\therefore\frac{H}{2} = V_{0}t - \frac{1}{2}gt^{2}$ …(i)
For body $B,\ u = 0,\ a = + g,\ S = \frac{H}{2}$
$\therefore\frac{H}{2} = \frac{1}{2}gt^{2}$ …(ii)
Equating equations (i) and (ii), we get
$V_{0}t - \frac{1}{2}gt^{2} = \frac{1}{2}gt^{2} \Rightarrow V_{0}t = gt^{2}$ or $t = \frac{V_{0}}{g}$
Substituting the value of $t$ in equation (i), we get
$\frac{H}{2} = V_{0} \times \left( \frac{V_{0}}{g} \right) - \frac{1}{2}g\left( \frac{V_{0}}{g} \right)^{2} = \frac{V_{0}^{2}}{g} - \frac{1}{2}\frac{V_{0}^{2}}{g}$
$\frac{H}{2} = \frac{1}{2}\frac{V_{0}^{2}}{g}$ or $V_{0}^{2} = gH \Rightarrow V_{0} = \sqrt{gH}$