If four balls A,B,C,D are projected with same speed at angles of `15^(@),30^(@),45^(@) and 60^(@)` with the horizontal respectively, the two balls which will fall at the same place will be-

Options

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

Answer

To solve the problem of which two balls will fall at the same place when projected at different angles, we need to calculate the range of each ball. The range \( R \) for a projectile is given by the formula:

\[ R = \frac{U^2 \sin(2\theta)}{g} \]

where:
- \( U \) is the initial speed,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 1: Calculate the Range for Each Ball

1. **For Ball A (θ = 15°)**:
\[ R_A = \frac{U^2 \sin(2 \times 15°)}{g} = \frac{U^2 \sin(30°)}{g} \]
Since \( \sin(30°) = \frac{1}{2} \):
\[ R_A = \frac{U^2 \cdot \frac{1}{2}}{g} = \frac{U^2}{2g} \]

2. **For Ball B (θ = 30°)**:
\[ R_B = \frac{U^2 \sin(2 \times 30°)}{g} = \frac{U^2 \sin(60°)}{g} \]
Since \( \sin(60°) = \frac{\sqrt{3}}{2} \):
\[ R_B = \frac{U^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3} U^2}{2g} \]

3. **For Ball C (θ = 45°)**:
\[ R_C = \frac{U^2 \sin(2 \times 45°)}{g} = \frac{U^2 \sin(90°)}{g} \]
Since \( \sin(90°) = 1 \):
\[ R_C = \frac{U^2}{g} \]

4. **For Ball D (θ = 60°)**:
\[ R_D = \frac{U^2 \sin(2 \times 60°)}{g} = \frac{U^2 \sin(120°)}{g} \]
Since \( \sin(120°) = \sin(90° + 30°) = \cos(30°) = \frac{\sqrt{3}}{2} \):
\[ R_D = \frac{U^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3} U^2}{2g} \]

### Step 2: Compare Ranges

Now we have the ranges:
- \( R_A = \frac{U^2}{2g} \)
- \( R_B = \frac{\sqrt{3} U^2}{2g} \)
- \( R_C = \frac{U^2}{g} \)
- \( R_D = \frac{\sqrt{3} U^2}{2g} \)

From the calculations, we see that:
- \( R_B = R_D \)

### Conclusion

The two balls that will fall at the same place are **Ball B (30°)** and **Ball D (60°)**.

---