Question

A body is projected from ground with a velocity ‘u’ making an angle of 60^o with horizontal. If the body just clears the building in a time interval of 2 sec, maximum height attained by the body is

Moderate

Solution

$u \cos α \times Δ t = l \Rightarrow u \cos 60^{o} \times 2 = 40 \Rightarrow u = 40 m / s$
$∴ H_{\max} = \frac{u^{2} \sin^{2} α}{2 g} = \frac{40^{2} \sin^{2} 60^{o}}{2 \times 10} m = 60 m$

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