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Question

Physics Question on Motion in a plane

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 π/3\pi / 3 and the maximum height reached by it is 102m102\, m. Then the maximum height reached by the other in metre is

A

336

B

224

C

56

D

34

Answer

34

Explanation

Solution

Key Idea Horizontal ranges are same for complementary angles of projection ie, for θ\theta and (90θ)\left(90^{\circ}-\theta\right)
We know that if two stones have same horizontal range, then this implies that both are projected at θ\theta and 90θ90^{\circ}-\theta
Here, θ=π3=60\theta=\frac{\pi}{3}=60^{\circ}
90θ=9060=30\therefore 90^{\circ}-\theta=90^{\circ}-60^{\circ}=30^{\circ}
For first stone, Max. height =102=u2sin2602g=102=\frac{u^{2} \sin ^{2} 60^{\circ}}{2 g}
For second stone, Max. height, h=u2sin2302gh=\frac{u^{2} \sin ^{2} 30^{\circ}}{2 g}
h102=sin230sin260=(1/2)2(3/2)2\therefore \frac{h}{102}=\frac{\sin ^{2} 30^{\circ}}{\sin ^{2} 60^{\circ}}=\frac{(1 / 2)^{2}}{(\sqrt{3} / 2)^{2}}
or h=102×1/43/4=34mh=102 \times \frac{1 / 4}{3 / 4}=34 \,m