A body is projected from the ground at an angle of `45^(@)` with the horizontal. Its velocity after 2s is 20 ms. The maximum height reached by the body during its motion is m. (use g = `g = 10 ms^(-2)`)

Explanation

To solve this problem, we need to determine the maximum height reached by a projectile launched at an angle of \(45^\circ\) with an initial velocity \(U\). Given that the velocity of the projectile after 2 seconds is \(20 \, \text{m/s}\) and \(g = 10 \, \text{m/s}^2\), we can follow these steps:

1. **Resolve the initial velocity into components:**
Since the projectile is launched at \(45^\circ\),
\[ U_x = U \cos 45^\circ = \frac{U}{\sqrt{2}} \]
\[
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