Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is `pi/3` and the maximum height reached by it is 102 m. Then the maximum height reached by the other in metres is

Options

A. 76
B. 84
C. 56
D. 34

Explanation

To solve the problem, we need to find the maximum height reached by the second stone when it is projected at an angle of \( \frac{\pi}{6} \) (or 30 degrees), given that the first stone is projected at \( \frac{\pi}{3} \) (or 60 degrees) and reaches a maximum height of 102 m.

### Step-by-Step Solution:

1. **Understanding the relationship between angles and ranges**:
Since the horizontal ranges of both stones are equal, they must be projected at complementary angles. If one stone is projected at \( \theta_1 = 60^\circ \), the other stone is projected at \( \theta_2 = 30^\circ \).

2. **Using the formula for maximum height**:
The formula for the maximum height \( H \) reached by a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial speed, \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)), and \( \theta \) is the angle of projection.

3. **Finding the initial speed \( u \)**:
For the first stone (projected at \( 60^\circ \)):
\[ H_1 = \frac{u^2 \sin^2(60^\circ)}{2g} \]
Given \( H_1 = 102 \, \text{m} \) and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ 102 = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2 \cdot 10} \]
\[ 102 = \frac{u^2 \cdot \frac{3}{4}}{20} \]
\[ 102 = \frac{3u^2}{80} \]
\[ 3u^2 = 102 \cdot 80 \]
\[ 3u^2 = 8160 \]
\[ u^2 = \frac{8160}{3} = 2720 \]
\[ u = \sqrt{2720} \approx 52.2 \, \text{m/s} \]

4. **Calculating the maximum height for the second stone**:
For the second stone (projected at \( 30^\circ \)):
\[ H_2 = \frac{u^2 \sin^2(30^\circ)}{2g} \]
Knowing \( \sin(30^\circ) = \frac{1}{2} \):
\[ H_2 = \frac{u^2 \left(\frac{1}{2}\right)^2}{2 \cdot 10} \]
\[ H_2 = \frac{u^2 \cdot \frac{1}{4}}{20} \]
\[ H_2 = \frac{u^2}{80} \]
Substituting \( u^2 = 2720 \):
\[ H_2 = \frac{2720}{80} = 34 \, \text{m} \]

### Final Answer:
The maximum height reached by the second stone is \( \boxed{34} \, \text{m} \).