A fighter plane moving with a speed of `50 sqrt(2) m s^-1` upward at an angle of `45^@` with the vertical releases a bomb when it was at a height `1000 m` from ground. Find (a) the time of flight
(b) the maximum height of the bomb above ground.

Explanation

To solve the problem step by step, we will break it down into two parts: (a) finding the time of flight and (b) finding the maximum height of the bomb above the ground.

### Given Data:
- Speed of the plane (u) = \(50 \sqrt{2} \, \text{m/s}\)
- Angle with the vertical = \(45^\circ\)
- Height from the ground (h_initial) = \(1000 \, \text{m}\)
- Acceleration due to gravity (g) = \(10 \, \text{m/s}^2\)

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