Kinematics

JEE Main level, numerical answer, medium

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Question 235 of 308

An object is projected in the air with initial velocity uu at an angle θ\theta. The projectile motion is such that the horizontal range RR, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be _____ degree.

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Answer: 15

The maximum range is given by, Rmax⁡=u2sin⁡2(45∘)g=u2g{R}_{\max }=\dfrac{{u}^{2}\sin 2\left(45^{\circ}\right)}{g}=\dfrac{{u}^{2}}{g}.
For the range to be half of the maximum range,
Rmax2=u22g=u2sin⁡2θg\dfrac{{R}_{max}}{2}=\dfrac{{u}^{2}}{2g}=\dfrac{{u}^{2}\sin 2\theta }{g}
⇒sin⁡2θ=12\Rightarrow \sin 2\theta =\dfrac{1}{2}
⇒2θ=30∘or2θ=150∘\Rightarrow 2\theta =30^{\circ}\text{or}2\theta =150^{\circ}
⇒θ=15∘,75∘\Rightarrow \theta =15^{\circ},75^{\circ}
Therefore, answer is 15or7515\text{or}75

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