Projectile motion is a combination of two one-dimensional motion: one in horizontal and other in vertical direction. Motion in 2D means in a plane. Necessary condition for 2D motion is that the velocity vector is coplanar to the acceleration vector. In case of projectile motion, the angle between velocity and acceleration will be `0^@ltthetalt180^@`. During the projectile motion, the horizontal component of velocity ramains unchanged but the vertical component of velocity is time dependent. Now answer the following questions:
A body is projected at angle of `30^@` and `60^@` with the same velocity. Their horizontal ranges are `R_1` and `R_2` and maximum heights are `H_1 and H_2`, respectively, then
A body is projected at angle of `30^@` and `60^@` with the same velocity. Their horizontal ranges are `R_1` and `R_2` and maximum heights are `H_1` and `H_2`, respectively, then

Options

A. `R_1/R_2gt1`
B. `H_1/H_2gt1`
C. `R_1/R_2lt1`
D. `H_1/H_2lt1`

Explanation

To solve the problem regarding the projectile motion of a body projected at angles of \(30^\circ\) and \(60^\circ\) with the same initial velocity, we need to determine the horizontal ranges \(R_1\) and \(R_2\) and the maximum heights \(H_1\) and \(H_2\) for both angles.

### Step 1: Calculate the Horizontal Ranges

The formula for the horizontal range \(R\) of a projectile is given by:

\[ R = \frac{U^2 \sin(2\theta)}{g} \]

Where:
- \(U\) is the initial velocity,
- \(\theta\) is the angle of projection,
- \(g\) is the acceleration due to gravity.

For the angle of \(30^\circ\):

\[ R_1 = \frac{U^2 \sin(2 \times 30^\circ)}{g} = \frac{U^2 \sin(60^\circ)}{g} \]

For the angle of \(60^\circ\):

\[ R_2 = \frac{U^2 \sin(2 \times 60^\circ)}{g} = \frac{U^2 \sin(120^\circ)}{g} \]

Since \(\sin(120^\circ) = \sin(60^\circ)\), we have:

\[ R_2 = \frac{U^2 \sin(60^\circ)}{g} \]

Now, comparing \(R_1\) and \(R_2\):

\[ R_1 = R_2 \]

Thus, the ratio of the ranges is:

\[ \frac{R_1}{R_2} = 1 \]

### Step 2: Calculate the Maximum Heights

The formula for the maximum height \(H\) of a projectile is given by:

\[ H = \frac{U^2 \sin^2(\theta)}{2g} \]

For the angle of \(30^\circ\):

\[ H_1 = \frac{U^2 \sin^2(30^\circ)}{2g} = \frac{U^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{U^2 \cdot \frac{1}{4}}{2g} = \frac{U^2}{8g} \]

For the angle of \(60^\circ\):

\[ H_2 = \frac{U^2 \sin^2(60^\circ)}{2g} = \frac{U^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{U^2 \cdot \frac{3}{4}}{2g} = \frac{3U^2}{8g} \]

Now, comparing \(H_1\) and \(H_2\):

\[ \frac{H_1}{H_2} = \frac{\frac{U^2}{8g}}{\frac{3U^2}{8g}} = \frac{1}{3} \]

### Conclusion

From the calculations:
- The ratio of the ranges \( \frac{R_1}{R_2} = 1 \)
- The ratio of the heights \( \frac{H_1}{H_2} = \frac{1}{3} \)

Thus, the correct options are:
- \(R_1/R_2 = 1\) (not provided in options)
- \(H_1/H_2 < 1\) (Option D)