(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 can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then their product is:

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$A$ particle is projected upwards with a velocity of $100 \ m/s$ at an angle of $60^{\circ}$ with the vertical. Find the time when the particle will move perpendicular to its initial direction,$.....$ seconds (take $g = 10 \ m/s^2$):-

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The range of a particle when launched at an angle of $15^\circ$ with the horizontal is $1.5 \, km$. What is the range of the projectile when launched at an angle of $45^\circ$ to the horizontal? (in $km$)

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An aeroplane is flying at a constant horizontal velocity of $600\, km/hr$ at an elevation of $6\, km$ towards a point directly above the target on the earth's surface. At an appropriate time,the pilot releases a ball so that it strikes the target at the earth. The ball will appear to be falling

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$A$ projectile is thrown with a velocity of $50 \, m/s$ at an angle of $53^o$ with the horizontal. Determine the instants at which the projectile is at the same height.

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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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$A$ projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then their product is:

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