A gun fires two bullets at `60^(@)` and `30^(@)` with horizontal. The bullets strike at same horizontal distance. The ratio of the maximum height for the two bullets is in the ratio

Options

A. `2 : 1`
B. `3 : 1`
C. `4 : 1`
D. `1 : 1`

Explanation

To solve the problem of finding the ratio of the maximum heights of two bullets fired at angles of \(60^\circ\) and \(30^\circ\) with the same initial speed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for maximum height**:
The maximum height \(h\) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where \(u\) is the initial speed, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

2. **Calculate the maximum height for the bullet fired at \(60^\circ\)**:
For the first bullet fired at an angle of \(60^\circ\):
\[ h_1 = \frac{u^2 \sin^2 60^\circ}{2g} \]
We know that \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), so:
\[ h_1 = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{3}{4}}{2g} = \frac{3u^2}{8g} \]

3. **Calculate the maximum height for the bullet fired at \(30^\circ\)**:
For the second bullet fired at an angle of \(30^\circ\):
\[ h_2 = \frac{u^2 \sin^2 30^\circ}{2g} \]
We know that \(\sin 30^\circ = \frac{1}{2}\), so:
\[ h_2 = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g} \]

4. **Find the ratio of the maximum heights**:
Now we need to find the ratio \(\frac{h_1}{h_2}\):
\[ \frac{h_1}{h_2} = \frac{\frac{3u^2}{8g}}{\frac{u^2}{8g}} = \frac{3u^2}{8g} \cdot \frac{8g}{u^2} = 3 \]

5. **Conclusion**:
The ratio of the maximum heights of the two bullets is:
\[ h_1 : h_2 = 3 : 1 \]

### Final Answer:
The ratio of the maximum height for the two bullets is \(3 : 1\).