(C) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.For maximum range,$\theta = 45^{\circ}$,so $R_{\max} = \frac{u^2}{g}$.For th

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(C) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.For maximum range,$\theta = 45^{\circ}$,so $R_{\max} = \frac{u^2}{g}$.For the second object,the range is $R' = \frac{R_{\max}}{2} = \frac{u^2}{2g}$.Equating the range formula for the second object: $\frac{u^2 \sin 2\theta'}{g} = \frac{u^2}{2g}$.This simplifies to $\sin 2\theta' = \frac{1}{2}$.Therefore,$2\theta' = 30^{\circ}$ or $2\theta' = 150^{\circ}$.Solving for $\theta'$,we get $\theta' = 15^{\circ}$ or $\theta' = 75^{\circ}$.

Class 11 Physics 2 Motion in a Plane Horizontal Projectile Motion

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An object is projected in the air with initial velocity $u$ at an angle $\theta$. The projectile motion is such that the horizontal range $R$ is maximum. Another object is projected in the air with a horizontal range half of the range of the first object. The initial velocity remains the same in both cases. The value of the angle of projection,at which the second object is projected,will be $.......$ degrees.

JEE MAIN 2022 All JEE MAIN

A
$85$
B
$80$
C
$15$ or $75$
D
$70$

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2 Motion in a Plane

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Horizontal Projectile Motion

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$A$ stone is projected horizontally with a speed $10 \ m/s$ from an $80 \ m$ high building. The distance of the target on the ground from the foot of the building is $.... \ m$ $(g = 10 \ m/s^2)$.

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