Question

If R is the range on a horizontal plane and T the time of flight of a projectile, then the angle of projection $\alpha$ given by $\tan \alpha$ is equal to?

• A. $g{T}^2/2R$
• B. $gT/2R$
• C. $T^2/Rg$
• D. $T^2/2Rg$

Answer 1

The correct option is A $g{T}^2/2R$

$\text{Step 1: Expression for Range and Time of flight of Projectile}$ $\text{[Ref. Fig.]}$

Let $u$ b the initial velocity of the projectile.

Range is given by: $R=\frac{u^2\sin 2\alpha }{g}$ $....\left(1\right)$

Time of flight is given by: $T=\frac{2u\sin \alpha }{g}$ $....\left(2\right)$

$\text{Step 2:}$ $\tan \alpha$ $\text{Calculation}$

From equation $\left(1\right)$ and $\left(2\right)$

$\frac{T^2}{R}=\frac{\left(\frac{4u^2{\sin }^2\alpha }{g^2}\right)}{\left(\frac{u^{2}2\sin \alpha \cos \alpha }{g}\right)}$$=\frac{2\tan \alpha }{g}$

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

Hence, Option $A$ is correct.