Two stones are projected so as to reach the same distance from the point of projection on a horizontal surface. The maximum height reached by one exceeds thr other by an amount equal to half the sum of the height attained by them. Then angle of projection of the stone which attains smaller height is

Options

A. `45^(@)`
B. `60^(@)`
C. `30^(@)`
D. `tan^(-1)(3//4)`

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the Problem
We have two stones projected at angles θ1 and θ2 such that they reach the same horizontal distance (range). We know that the maximum height of one stone exceeds the other by an amount equal to half the sum of their heights.

### Step 2: Establish Relationships
Since both stones reach the same range, their angles of projection are complementary:
- If θ1 is the angle of projection for the first stone, then θ2 = 90° - θ1 for the second stone.

### Step 3: Set Up the Height Equation
Let h1 be the maximum height of the first stone and h2 be the maximum height of the second stone. According to the problem:
\[ h1 - h2 = \frac{1}{2}(h1 + h2) \]

### Step 4: Simplify the Height Equation
Rearranging the above equation gives:
\[ 2(h1 - h2) = h1 + h2 \]
This simplifies to:
\[ 2h1 - 2h2 = h1 + h2 \]
\[ 2h1 - h1 = 2h2 + h2 \]
\[ h1 = 3h2 \]

### Step 5: Use the Formula for Maximum Height
The maximum height for projectile motion is given by:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
Thus, we can express h1 and h2 as:
- \( h1 = \frac{u^2 \sin^2 \theta_1}{2g} \)
- \( h2 = \frac{u^2 \sin^2 \theta_2}{2g} \)

### Step 6: Substitute for h1 and h2
Substituting \( h1 \) and \( h2 \) into the equation \( h1 = 3h2 \):
\[ \frac{u^2 \sin^2 \theta_1}{2g} = 3 \left( \frac{u^2 \sin^2 \theta_2}{2g} \right) \]

### Step 7: Cancel Common Terms
Since \( \frac{u^2}{2g} \) is common on both sides, we can cancel it out:
\[ \sin^2 \theta_1 = 3 \sin^2 \theta_2 \]

### Step 8: Use the Complementary Angle Identity
Using the identity \( \sin(90° - \theta) = \cos \theta \), we have:
\[ \sin^2 \theta_2 = \cos^2 \theta_1 \]
Thus, substituting gives:
\[ \sin^2 \theta_1 = 3 \cos^2 \theta_1 \]

### Step 9: Use the Pythagorean Identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
Let \( \sin^2 \theta_1 = x \), then \( \cos^2 \theta_1 = 1 - x \):
\[ x = 3(1 - x) \]
\[ x = 3 - 3x \]
\[ 4x = 3 \]
\[ x = \frac{3}{4} \]
Thus, \( \sin^2 \theta_1 = \frac{3}{4} \) and \( \cos^2 \theta_1 = \frac{1}{4} \).

### Step 10: Find θ1
Taking the square root:
\[ \sin \theta_1 = \frac{\sqrt{3}}{2} \]
This corresponds to:
\[ \theta_1 = 60° \]

### Step 11: Find θ2
Since \( \theta_2 = 90° - \theta_1 \):
\[ \theta_2 = 90° - 60° = 30° \]

### Final Answer
The angle of projection of the stone which attains the smaller height is **30°**.

---