Question

A body is projected from a point on ground with velocity 40 m/s. If the velocity of projection and the velocity with which it strikes the ground are mutually perpendicular, then maximum height attained by the body is

Moderate

Solution

$α + α = 90^{o} \Rightarrow α = 45^{o}$
$\begin{array}{l} ∴ H_{\max} = \frac{u^{2} \sin^{2} α}{2 g} = \frac{{(40)}^{2} \sin^{2} 45^{o}}{2 \times 10} m \\ = 40 m \end{array}$

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