Four bodies P, Q, R and S are projected with equal velocities having angles of projection 15°, 30°, 45° and 60° with the horizontal respectively. The body having shortest range is

The correct option is P.

To find which body has the shortest horizontal range, we use the formula for the horizontal range of a projectile:
$R=\frac{u^2\sin \left(2\theta \right)}{g}$
where:
- $u$ is the velocity of projection,
- $\theta$ is the angle of projection with the horizontal,
- $g$ is the acceleration due to gravity.

Since all four bodies are projected with equal velocities ($u$ is constant) and under the same gravitational acceleration ($g$ is constant), the range $R$ is directly proportional to $\sin \left(2\theta \right)$:
$R\propto \sin \left(2\theta \right)$

Let us calculate the value of $\sin \left(2\theta \right)$ for each of the given angles:
1. For body P ($\theta ={15}^{\circ }$):
$\sin (2\times {15}^{\circ })=\sin \left({30}^{\circ }\right)=0.5$
2. For body Q ($\theta ={30}^{\circ }$):
$\sin (2\times {30}^{\circ })=\sin \left({60}^{\circ }\right)=\frac{\sqrt{3}}{2}\approx 0.866$
3. For body R ($\theta ={45}^{\circ }$):
$\sin (2\times {45}^{\circ })=\sin \left({90}^{\circ }\right)=1$ (Maximum Range)
4. For body S ($\theta ={60}^{\circ }$):
$\sin (2\times {60}^{\circ })=\sin \left({120}^{\circ }\right)=\sin ({180}^{\circ }-{60}^{\circ })=\sin \left({60}^{\circ }\right)\approx 0.866$

Comparing the values of $\sin \left(2\theta \right)$, the minimum value is $0.5$, which corresponds to body P projected at ${15}^{\circ }$. Therefore, body P has the shortest range.