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

Updated On: Jul 8, 2024
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The Correct Option is D

Solution and Explanation

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

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration