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. `(u^(2))/(g)`
B. `(2u^(2))/(g)`
C. `(u^(2))/(2g)`
D. `(u^(2))/(4g)`

Explanation

To solve the problem, we need to find the sum of the maximum heights attained by a ball thrown at two different angles but with the same initial speed and range. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a ball thrown at two different angles, \( \theta_1 \) and \( \theta_2 \), with the same initial speed \( u \).
- The range of the projectile is the same for both angles, which implies that \( \theta_1 + \theta_2 = 90^\circ \) (complementary angles).

2. **Formula for Maximum Height**:
- The maximum height \( h \) attained by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
- Here, \( g \) is the acceleration due to gravity.

3. **Calculating Heights**:
- For the first angle \( \theta_1 \):
\[ y_1 = \frac{u^2 \sin^2 \theta_1}{2g} \]
- For the second angle \( \theta_2 \):
\[ y_2 = \frac{u^2 \sin^2 \theta_2}{2g} \]

4. **Using the Complementary Angle Property**:
- Since \( \theta_1 + \theta_2 = 90^\circ \), we can express \( \theta_2 \) as:
\[ \theta_2 = 90^\circ - \theta_1 \]
- Therefore, we can use the identity \( \sin(90^\circ - \theta) = \cos(\theta) \):
\[ y_2 = \frac{u^2 \sin^2 (90^\circ - \theta_1)}{2g} = \frac{u^2 \cos^2 \theta_1}{2g} \]

5. **Summing the Heights**:
- Now, we can find \( y_1 + y_2 \):
\[ y_1 + y_2 = \frac{u^2 \sin^2 \theta_1}{2g} + \frac{u^2 \cos^2 \theta_1}{2g} \]
- Factor out \( \frac{u^2}{2g} \):
\[ y_1 + y_2 = \frac{u^2}{2g} (\sin^2 \theta_1 + \cos^2 \theta_1) \]

6. **Using the Pythagorean Identity**:
- We know that \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ y_1 + y_2 = \frac{u^2}{2g} \cdot 1 = \frac{u^2}{2g} \]

### Final Result:
Thus, the value of \( y_1 + y_2 \) is:
\[ y_1 + y_2 = \frac{u^2}{2g} \]