A ball is thrown vertically upwards from the ground G with a speed u. It reaches a point B at a height h (lower than the maximum height) after time t₁ It returns to the ground after time f, from the instant it was at B during the upward journey. Then

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$t_{1} t_{2} = (2 h / g)$

$t_{1} t_{2} = (h / g)$

$t_{1} t_{2} = (h / 2 g)$

$t_{1} t_{2} = (h / 4 g)$

answer is A.

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Detailed Solution

Here, $t = t_{1} + t_{2} = (2 u / g)$
$∴ u = \frac{(t_{1} + t_{2}) g}{2}$
Now $h = {ut}_{1} - \frac{1}{2} {gt}_{1}^{2}$
Or $h = \frac{(t_{1} + t_{2}) {gt}_{1}}{2} - \frac{1}{2} {gt}_{1}^{2}$
$\begin{array}{l} = \frac{1}{2} {gt}_{1}^{2} + \frac{1}{2} {gt}_{1} t_{2} - \frac{1}{2} {gt}_{1}^{2} \\ = \frac{1}{2} {gt}_{1} t_{2} \\ ∴ t_{1} t = (2 h / g) \end{array}$