First slide

Projection Under uniform Acceleration

Question

A body is projected at an angle $60^{0}$ with the horizontal and it just crosses a wall of height $\sqrt{3}$ m and hits the ground 2m behind the wall. Then the distance of the wall from the point of projection is

Moderate

A

$3 m$
B

$4 m$
C

$1.5 m$
D

$2 m$

Solution

If height of the wall is y, then $y = x \tan θ (1 - \frac{x}{R})$ x is the distance of the wall from the point of projection
$y = x \tan θ (1 - \frac{x}{x + 2})$ here $θ$ = $60^{0}$

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