(B) The horizontal range $R$ is given by $R = u \cos \theta \cdot T$,where $u$ is the initial velocity and $\theta$ is the angle of projection.The tim

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$\tan^{-1} \left( \frac{gT^2}{2R} \right)$

(B) The horizontal range $R$ is given by $R = u \cos \theta \cdot T$,where $u$ is the initial velocity and $\theta$ is the angle of projection.The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$,which implies $u \sin \theta = \frac{gT}{2}$.Dividing the vertical component by the horizontal component:$\tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{gT/2}{R/T} = \frac{gT^2}{2R}$.Therefore,the angle of projection is $\theta = \tan^{-1} \left( \frac{gT^2}{2R} \right)$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

Medium

If the time of flight of a bullet over a horizontal range $R$ is $T \ s$,the inclination of the direction of projection to the horizontal is

A
$\sin^{-1} \left( \frac{gT^2}{R} \right)$
B
$\tan^{-1} \left( \frac{gT^2}{2R} \right)$
C
$\cos^{-1} \left( \frac{2gT^2}{2R} \right)$
D
$\cot^{-1} \left( \frac{R}{gT^2} \right)$

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3-2.Motion in Plane

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Horizontal Projectile Motion

$A$ projectile is thrown at an angle of $\pi/3$ with the horizontal,reaching a maximum height $Y$. If another projectile is thrown with the same initial velocity at an angle of $\pi/6$,what will be the maximum height attained?

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An object is thrown along a direction inclined at an angle of $45^\circ$ with the horizontal direction. The horizontal range of the particle is equal to

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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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$A$ boy weighing $50 \ kg$ finished a long jump at a distance of $8 \ m$. Considering that he moved along a parabolic path and his angle of jump is $45^{\circ}$,his initial $K.E.$ is (in $J$)

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$A$ particle is thrown with a speed of $12 \, m/s$ at an angle $60^o$ with the horizontal. The time interval between the moments when its speed is $10 \, m/s$ is $(g = 10 \, m/s^2)$.........$s$

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If the time of flight of a bullet over a horizontal range $R$ is $T \ s$,the inclination of the direction of projection to the horizontal is

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