Question

A projectile is projected with some velocity at an angle $\theta$ from the horizontal so that its range is $R$ and time of flight is $T$ then

• A. $R=\frac{g^2{T}^2}{\sqrt{2}\tan \theta }$
• B. $R=\frac{g{T}^{\frac{1}{2}}}{3\tan \theta }$
• C. $R=\frac{g{T}^2}{2\tan \theta }$
• D. $R=\frac{\sqrt{3}g{T}^3}{3\tan \theta }$

Answer 1

The correct option is C $R=\frac{g{T}^2}{2\tan \theta }$

Lets consider, $u=$ initial velocity of the projectile

$\theta =$ angle of projection

Time of flight in projectile motion,

$T=\frac{2usin\theta }{g}$

$usin\theta =\frac{gT}{2}$. . . . . . . . . . . (1)

The Horizontal range in projectile motion,

$R=\frac{2u^{2}sin\theta cos\theta }{g}$

$R=\frac{2(usin\theta {)}^{2}cos\theta }{gsin\theta }$

$R=\frac{2\times (\frac{gT}{2}{)}^2}{gtan\theta }$

$R=\frac{g{T}^2}{2tan\theta }$

The correct option is C.