Two bodies are projected from ground with equal speeds 20 m/sec from the same position in same vertical plane to have equal range but at different angle above the horizontal. If one of the angle is 30° the sum of their maximum heights is (assume `g = 10m//s^2`)

Options

A. 400 m
B. 20 m
C. 30 m
D. 40 m

Explanation

To solve the problem, we need to find the sum of the maximum heights of two bodies projected at different angles but with the same initial speed and equal range. One of the angles is given as 30°. We will also use the fact that the angles that give the same range are complementary angles.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Initial speed (u) = 20 m/s
- Angle 1 (θ₁) = 30°
- Angle 2 (θ₂) = 90° - θ₁ = 90° - 30° = 60°
- Acceleration due to gravity (g) = 10 m/s²

2. **Formula for Maximum Height:**
The maximum height (H) reached by a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

3. **Calculate Maximum Height for θ₁ (30°):**
- Substitute θ₁ into the formula:
\[ H₁ = \frac{(20)^2 \sin^2(30°)}{2 \cdot 10} \]
- We know that \(\sin(30°) = \frac{1}{2}\), so:
\[ H₁ = \frac{400 \cdot \left(\frac{1}{2}\right)^2}{20} = \frac{400 \cdot \frac{1}{4}}{20} = \frac{100}{20} = 5 \text{ m} \]

4. **Calculate Maximum Height for θ₂ (60°):**
- Substitute θ₂ into the formula:
\[ H₂ = \frac{(20)^2 \sin^2(60°)}{2 \cdot 10} \]
- We know that \(\sin(60°) = \frac{\sqrt{3}}{2}\), so:
\[ H₂ = \frac{400 \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{20} = \frac{400 \cdot \frac{3}{4}}{20} = \frac{300}{20} = 15 \text{ m} \]

5. **Sum of Maximum Heights:**
- Now, we add the two maximum heights:
\[ H_{total} = H₁ + H₂ = 5 \text{ m} + 15 \text{ m} = 20 \text{ m} \]

### Final Answer:
The sum of their maximum heights is **20 meters**.