Four bodies A, B, C and D are projected with equal speeds having angles of projection 15${}^{\circ }$, 30${}^{\circ }$, 45${}^{\circ }$ and 60${}^{\circ }$ with the horizontal respectively. The body having the shortest range is:

The correct option is A A

$\text{Step 1: Range of the projectile}$

Let $u$ be the equal speed of all four bodies.

Let $\theta$ be the angle of projection.

$R=\frac{u^2\sin 2\theta }{g}$ $....\left(1\right)$

$\text{Step 2: Range of all the bodies}$

Substitute the values of $\theta$ in equation $\left(1\right)$ for different bodies

Range of A, $R_1=\frac{u^2\sin {30}^o}{g}=\frac{u^2}{2g}$

Range of B, $R_2=\frac{u^2\sin {60}^o}{g}=\frac{\sqrt{3}{u}^2}{2g}$

Range of C, $R_3=\frac{u^2\sin {90}^o}{g}=\frac{u^2}{g}$

Range of D, $R_4=\frac{u^2\sin {120}^o}{g}=\frac{\sqrt{3}{u}^2}{2g}$

Range of A is the shortest

Hence, option $\left(A\right)$ is correct.

$\text{Note}$ - Range is maximum when $\theta ={45}^o$