Two particle are projected with same initial velocities at an angle `30^(@)` and `60^(@)` with the horizontal .Then

Options

A. their heights will be equal
B. their height will be different
C. their range of flight will be equal
D. their ranges will be different

Answer

To solve the problem, we need to analyze the projectile motion of two particles projected at angles of \(30^\circ\) and \(60^\circ\) with the same initial velocity. We will calculate the maximum heights and ranges for both projectiles.

### Step 1: Calculate the maximum height for both projectiles

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

\[ h = \frac{u^2 \sin^2 \theta}{2g} \]

Where:
- \(u\) = initial velocity
- \(\theta\) = angle of projection
- \(g\) = acceleration due to gravity

Let's denote:
- For the first projectile (angle \(30^\circ\)): \(h_1 = \frac{u^2 \sin^2 30^\circ}{2g}\)
- For the second projectile (angle \(60^\circ\)): \(h_2 = \frac{u^2 \sin^2 60^\circ}{2g}\)

Now, we know:
- \(\sin 30^\circ = \frac{1}{2}\)
- \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)

Calculating \(h_1\) and \(h_2\):

\[ h_1 = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g} \]

\[ h_2 = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{3}{4}}{2g} = \frac{3u^2}{8g} \]

### Step 2: Find the ratio of maximum heights

Now, we can find the ratio of the maximum heights:

\[ \frac{h_1}{h_2} = \frac{\frac{u^2}{8g}}{\frac{3u^2}{8g}} = \frac{1}{3} \]

This means:

\[ h_1 = \frac{1}{3} h_2 \]

### Step 3: Calculate the range for both projectiles

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

\[ R = \frac{u^2 \sin 2\theta}{g} \]

For the first projectile (angle \(30^\circ\)):

\[ R_1 = \frac{u^2 \sin 60^\circ}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]

For the second projectile (angle \(60^\circ\)):

\[ R_2 = \frac{u^2 \sin 120^\circ}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]

### Step 4: Find the ratio of ranges

Now, we can find the ratio of the ranges:

\[ \frac{R_1}{R_2} = \frac{\frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g}}{\frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g}} = 1 \]

This means:

\[ R_1 = R_2 \]

### Conclusion

- The maximum height of the projectile at \(30^\circ\) is one-third of the height at \(60^\circ\).
- The ranges of both projectiles are equal.

### Final Answers
1. \(h_1 = \frac{1}{3} h_2\) (Heights are different)
2. \(R_1 = R_2\) (Ranges are the same)