A parrachutist bails out from an aeropane and after dropping through a distance of `40 m` opens the parachute and decelerates at `2 ms^(-2)`. If he reaches the ground with a speed of `2 ms^(-1)`, how lowg he in the air ? At what height did he bail out from the plane ?

Explanation

To solve the problem, we will break it down into steps:

### Step 1: Calculate the velocity just before the parachutist opens the parachute
The parachutist falls freely under gravity for a distance of 40 m. We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( u = 0 \) (initial velocity)
- \( a = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
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