Two bodies are thrown from the same point with the same velocity of projection angles of projection being complimentary angles. If `R_(1) and R_(2)` are the ranges and `h_(1) and h_(2)` are maximum heights respectively, then

Options

A. `R_(1)=R_(2)=(u^(2))/(g) sin 2 theta`
B. `h_(1)+h_(2)=(u^(2))/(2g)`
C. both
D. none

Explanation

To solve the problem, we need to analyze the motion of two bodies thrown at complementary angles with the same initial velocity. Let’s denote the angles of projection as \( \theta \) and \( 90^\circ - \theta \).

### Step-by-Step Solution:

1. **Understanding the Ranges**:
- The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
- For the first body projected at angle \( \theta \):
\[ R_1 = \frac{u^2 \sin(2\theta)}{g} \]
- For the second body projected at angle \( 90^\circ - \theta \):
\[ R_2 = \frac{u^2 \sin(2(90^\circ - \theta))}{g} = \frac{u^2 \sin(180^\circ - 2\theta)}{g} = \frac{u^2 \sin(2\theta)}{g} \]
- Thus, we find that:
\[ R_1 = R_2 \]

2. **Understanding the Maximum Heights**:
- The maximum height \( H \) of a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]
- For the first body:
\[ H_1 = \frac{u^2 \sin^2(\theta)}{2g} \]
- For the second body:
\[ H_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2(\theta)}{2g} \]
- Therefore, we have:
\[ H_1 + H_2 = \frac{u^2 \sin^2(\theta)}{2g} + \frac{u^2 \cos^2(\theta)}{2g} = \frac{u^2 (\sin^2(\theta) + \cos^2(\theta))}{2g} = \frac{u^2}{2g} \]

3. **Conclusion**:
- From the calculations, we conclude:
- The ranges are equal: \( R_1 = R_2 \)
- The sum of the maximum heights is: \( H_1 + H_2 = \frac{u^2}{2g} \)

### Final Results:
- \( R_1 = R_2 \)
- \( H_1 + H_2 = \frac{u^2}{2g} \)