First slide

Projection Under uniform Acceleration

Question

Two projectiles are projected with the same velocity. If one is projected at an angle of 30^o and the other at 60^o to the horizontal, then ratio of maximum heights reached, is

Easy

A

3 : 1
B

1 : 3
C

1 : 2
D

2 : 1

Solution

$h_{max} = \frac{u^{2} \sin^{2} θ}{2 g} \propto \sin^{2} θ$
$\begin{array}{l} ∴ \frac{{(h_{1})}_{max}}{{(h_{2})}_{max}} = \frac{{(\sin θ_{1})}^{2}}{\sin {(θ_{2})}^{2}} = \frac{{(\sin 30^{\circ})}^{2}}{{(\sin 60^{\circ})}^{2}} \\ = \frac{(1 / 2)^{2}}{(\sqrt{3} / 2)^{2}} = \frac{1}{3} \end{array}$
or ${(h_{1})}_{max} : {(h_{2})}_{max} = 1 : 3$

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