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 find the maximum height reached by the second stone projected at an angle of \( \theta_2 \) with the same initial speed as the first stone, we can follow these steps:

### Step 1: Understand the Problem
We have two stones projected with the same speed but at different angles. The angle of projection for the first stone is \( \theta_1 = \frac{\pi}{3} \) and it reaches a maximum height \( h_1 = 102 \, \text{m} \). We need to find the maximum height \( h_2 \) of the second stone, which is projected at an angle \( \theta_2 \) such that the horizontal ranges of both stones are equal.

### Step 2: Use the Range Formula
The horizontal range \( R \) for a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Since the ranges are equal, we can set up the equation:
\[ \frac{u^2 \sin(2\theta_1)}{g} = \frac{u^2 \sin(2\theta_2)}{g} \]
This simplifies to:
\[ \sin(2\theta_1) = \sin(2\theta_2) \]

### Step 3: Find the Angle \( \theta_2 \)
Using the identity \( \sin(2\theta) = \sin(\pi - 2\theta) \), we can find \( \theta_2 \):
\[ 2\theta_1 + 2\theta_2 = \pi \]
Substituting \( \theta_1 = \frac{\pi}{3} \):
\[ 2 \cdot \frac{\pi}{3} + 2\theta_2 = \pi \]
\[ \frac{2\pi}{3} + 2\theta_2 = \pi \]
\[ 2\theta_2 = \pi - \frac{2\pi}{3} = \frac{\pi}{3} \]
\[ \theta_2 = \frac{\pi}{6} \]

### Step 4: Use the Height Formula
The maximum height \( h \) for a projectile is given by:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]
For the first stone:
\[ h_1 = \frac{u^2 \sin^2(\theta_1)}{2g} = 102 \]
For the second stone:
\[ h_2 = \frac{u^2 \sin^2(\theta_2)}{2g} \]

### Step 5: Relate the Heights
We can relate the heights of the two stones:
\[ \frac{h_1}{h_2} = \frac{\sin^2(\theta_1)}{\sin^2(\theta_2)} \]
Substituting the known values:
\[ \frac{102}{h_2} = \frac{\sin^2\left(\frac{\pi}{3}\right)}{\sin^2\left(\frac{\pi}{6}\right)} \]
Calculating the sine values:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Thus:
\[ \frac{102}{h_2} = \frac{\left(\frac{\sqrt{3}}{2}\right)^2}{\left(\frac{1}{2}\right)^2} = \frac{\frac{3}{4}}{\frac{1}{4}} = 3 \]
This gives us:
\[ h_2 = \frac{102}{3} = 34 \, \text{m} \]

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