A ball is thrown at different angles with the same speed `u` and from the same points and it has same range in both the cases. If `y_1 and y_2` be the heights attained in the two cases, then find the value of `y_1 + y_2`.

Options

A. (a)`(k^2)/g`
B. (b)`(2k^2)/g`
C. (c)`(k^2)/(2g)`
D. (d)`(k^2)/(4g)`

Explanation

To solve the problem, we need to find the sum of the maximum heights \( y_1 \) and \( y_2 \) attained by a ball thrown at different angles but with the same initial speed \( u \) and achieving the same range.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The ball is thrown at two different angles \( \theta \) and \( 90^\circ - \theta \) (complementary angles) with the same speed \( u \).
- The range \( R \) for both angles is the same.

2. **Formula for Range**:
- The range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
- Since the range is the same for both angles, we can write:
\[ R = \frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin(2(90^\circ - \theta))}{g} \]
- This confirms that the ranges are equal for complementary angles.

3. **Finding Maximum Heights**:
- The maximum height \( y \) attained by a projectile is given by:
\[ y = \frac{u^2 \sin^2(\theta)}{2g} \]
- For the angle \( \theta \), the height is:
\[ y_1 = \frac{u^2 \sin^2(\theta)}{2g} \]
- For the angle \( 90^\circ - \theta \), the height is:
\[ y_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2(\theta)}{2g} \]

4. **Sum of Heights**:
- Now, we can find the sum of the heights:
\[ y_1 + y_2 = \frac{u^2 \sin^2(\theta)}{2g} + \frac{u^2 \cos^2(\theta)}{2g} \]
- Factoring out \( \frac{u^2}{2g} \):
\[ y_1 + y_2 = \frac{u^2}{2g} (\sin^2(\theta) + \cos^2(\theta)) \]
- Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ y_1 + y_2 = \frac{u^2}{2g} \cdot 1 = \frac{u^2}{2g} \]

5. **Final Result**:
- Therefore, the value of \( y_1 + y_2 \) is:
\[ y_1 + y_2 = \frac{u^2}{2g} \]