If two bodies are projected at 30° and 60° respectively, with the same velocity, then

The correct option is Their ranges are same.

To understand why this is correct, we can analyze the formulas for the horizontal range, maximum height, and time of flight of a projectile.

1. Horizontal Range ($R$):
The horizontal range of a projectile launched with an initial velocity $u$ at an angle $\theta$ with the horizontal is given by the formula:

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

where $g$ is the acceleration due to gravity.

Let us calculate the range for both projection angles, keeping the initial velocity $u$ constant:
For the first body projected at ${\theta }_1={30}^{\circ }$:

$R_1=\frac{u^{2}sin(2\times {30}^{\circ })}{g}=\frac{u^{2}sin\left({60}^{\circ }\right)}{g}$

For the second body projected at ${\theta }_2={60}^{\circ }$:

$R_2=\frac{u^{2}sin(2\times {60}^{���})}{g}=\frac{u^{2}sin\left({120}^{\circ }\right)}{g}$

Using the trigonometric identity $sin({180}^{\circ }-A)=sin\left(A\right)$, we have:

$sin\left({120}^{\circ }\right)=sin({180}^{\circ }-{60}^{\circ })=sin\left({60}^{\circ }\right)$

Therefore, the ranges are equal:

$R_1=R_2$

In general, for two complementary angles of projection (where ${\theta }_1+{\theta }_2={90}^{\circ }$), the horizontal ranges are always identical when projected with the same initial velocity.

2. Why other options are incorrect:
The maximum height ($H$) is given by:

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

The time of flight ($T$) is given by:

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

Since $sin\left({30}^{\circ }\right)=0.5$ and $sin\left({60}^{\circ }\right)\approx 0.866$ are not equal, the maximum heights and the times of flight for both projectiles are different. Thus, only their ranges are the same.