Four bodies A,B,C and D are projected with equal velocities having angles of projection `15^(@),30^(@),45^(@) and 60^(@)` with the horizontal respectively. The body having the shortest range is

Options

A. A
B. B
C. C
D. D

Explanation

To determine which body has the shortest range when projected at different angles with the same initial velocity, we can use the formula for the range of a projectile:

\[ R = \frac{u^2 \sin 2\theta}{g} \]

where:
- \( R \) is the range,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

Since all bodies are projected with equal velocities, we can ignore \( u^2/g \) for our comparison and focus on the term \( \sin 2\theta \).

### Step 1: Calculate \( \sin 2\theta \) for each angle.
- For body A (15°):
\[ \sin 2\theta = \sin(2 \times 15°) = \sin 30° = \frac{1}{2} \]

- For body B (30°):
\[ \sin 2\theta = \sin(2 \times 30°) = \sin 60° = \frac{\sqrt{3}}{2} \approx 0.866 \]

- For body C (45°):
\[ \sin 2\theta = \sin(2 \times 45°) = \sin 90° = 1 \]

- For body D (60°):
\[ \sin 2\theta = \sin(2 \times 60°) = \sin 120° = \frac{\sqrt{3}}{2} \approx 0.866 \]

### Step 2: Compare the values of \( \sin 2\theta \).
- A (15°): \( \sin 2\theta = \frac{1}{2} \)
- B (30°): \( \sin 2\theta \approx 0.866 \)
- C (45°): \( \sin 2\theta = 1 \)
- D (60°): \( \sin 2\theta \approx 0.866 \)

### Step 3: Determine the smallest value.
From the calculations:
- The smallest value of \( \sin 2\theta \) is for body A (15°), which is \( \frac{1}{2} \).

### Conclusion:
Since the range \( R \) is directly proportional to \( \sin 2\theta \), the body with the smallest \( \sin 2\theta \) will have the shortest range. Therefore, the body with the shortest range is body A, which is projected at an angle of 15°.

### Final Answer:
The body having the shortest range is **A (15°)**.

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