Question

A body is projected with a velocity of 40 m/s such that its range is $4 / \sqrt{3} times$ the maximum height attained by it. If g = 10 m/s², the time of flight is

Moderate

Solution

$tanα= \frac{4H}{R} = \frac{4H}{4H/ \sqrt{3}} = \sqrt{3}$
$\Rightarrow α {=60}^{o}$
$∴ T = \frac{2 \times u \sin α}{g} = \frac{2 \times 40 \times \sin 60^{o}}{10} = 4 \sqrt{3} \sec$

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