Two stones are projected with the same speed but making different angles with the horizontal. Their ranges are equal. If the angles of projection of one is `pi//3` and its maximum height is `h_(1)` then the maximum height of the other will be:

Options

A. `3h_(1)`
B. `2h_(1)`
C. `h_(1)//2`
D. `h_(1)//3`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have two stones projected with the same speed but at different angles, and their ranges are equal. We know the angle of projection for the first stone is \( \theta_1 = \frac{\pi}{3} \) and we need to find the maximum height of the second stone, denoted as \( h_2 \).

### Step 2: Use the range formula
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial speed, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

Since the ranges of both stones 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: Determine the second angle
Given \( \theta_1 = \frac{\pi}{3} \):
\[ 2\theta_1 = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3} \]
Thus, we have:
\[ \sin\left(\frac{2\pi}{3}\right) = \sin(2\theta_2) \]
Using the identity \( \sin(\pi - x) = \sin(x) \), we can find \( \theta_2 \):
\[ 2\theta_2 = \frac{\pi}{3} \quad \text{or} \quad 2\theta_2 = \frac{2\pi}{3} \]
From \( 2\theta_2 = \frac{\pi}{3} \):
\[ \theta_2 = \frac{\pi}{6} \]

### Step 4: Calculate maximum heights
The maximum height \( h \) of a projectile is given by:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]
For the first stone:
\[ h_1 = \frac{u^2 \sin^2\left(\frac{\pi}{3}\right)}{2g} \]
Calculating \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\[ h_1 = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{3}{4}}{2g} = \frac{3u^2}{8g} \]

For the second stone:
\[ h_2 = \frac{u^2 \sin^2\left(\frac{\pi}{6}\right)}{2g} \]
Calculating \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \):
\[ h_2 = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g} \]

### Step 5: Relate \( h_2 \) to \( h_1 \)
Now, we can express \( h_2 \) in terms of \( h_1 \):
\[ h_2 = \frac{1}{3} h_1 \]
Substituting \( h_1 = \frac{3u^2}{8g} \):
\[ h_2 = \frac{1}{3} \cdot \frac{3u^2}{8g} = \frac{u^2}{8g} \]

### Final Answer
Thus, the maximum height of the second stone is:
\[ h_2 = \frac{u^2}{8g} \]