First slide

Projection Under uniform Acceleration

Question

Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is $\frac{π}{3}$ and the maximum height reached by it is 102 m. Then the maximum height reached by the other in meter is:

Moderate

A

336
B

224
C

56
D

34

Solution

$H_{\max} = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow 120 = \frac{u^{2} \sin^{2} π / 3}{2 g} \Rightarrow \frac{u^{2}}{2 g} = \frac{4}{3} (102)$
so, $H_{\max} = \frac{u^{2} \sin^{2} θ}{2 g}$
$= \frac{4}{3} (102) [\frac{1}{4}] = \frac{102}{3} = 34 m$

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