A projectile is thrown with speed u making angle `theta` with horizontal at `t =0`. It just crosses two points of equal height at time t = 1 s and t = 3 s respectively. Calculate the maximum height attained by it ? (g = `10 m//s^(2)`)

Explanation

To solve the problem of finding the maximum height attained by a projectile thrown with speed \( u \) at an angle \( \theta \) with the horizontal, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Problem**:
The projectile crosses two points of equal height at \( t = 1 \) s and \( t = 3 \) s. This indicates that the time taken to reach the maximum height is the midpoint of these two times. Therefore, the time to reach maximum height \( t_{max} \) is:
\[
t_{max} = \frac{1 + 3}{2} = 2 \text{ s}
...