Question

A particle is projected at an angle of 45° 2m from the foot of a wall, just touches the top of the wall and falls on the ground on the opposite side at a distance 4 m from it. The height of wall is :

Moderate

Solution

$Here θ = 45^{\circ} and \frac{u^{2}}{g} = 6 or u^{2} = 6 g, also ucos θt = 2$
$or t = \frac{2}{ucos θ}$
and y = height of wall, then
$y = usin θ \cdot t + \frac{1}{2} {gt}^{2} = usin θ \times \frac{2}{ucos θ} + \frac{1}{2} g \times \frac{4}{u^{2} \cos^{2} θ} or y = \frac{8}{3} m$

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