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PhysicsKinematics2-D & 3DEasy2 minPYQ_2022

PhysicsEasysingle choice

A ball is projected from the ground with a speed $15 {m s}^{- 1}$ at an angle $θ$ with horizontal so that its range and maximum height are equal, then $\tan θ$ will be equal to

Options:

Answer:

Solution:

The range and the maximum height of the projectile are given as below,
$R = \frac{u^{2} \sin 2 θ}{g} = \frac{2 u^{2} \sin θ \cos θ}{g} H = \frac{u^{2} \sin^{2} θ}{2 g}$
Therefore,
As $H = R$
$\Rightarrow \tan θ = 4$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: PYQ_2022

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