# A paratrooper jumps out of a stationary helicopter so that his initial velocity is 2ms–1 vertically downwards. He falls freely under gravity for 1·5 s, then his parachute opens and he descends...

A paratrooper jumps out of a stationary helicopter so that his initial velocity is 2ms–1 vertically downwards. He falls freely under gravity for 1·5 s, then his parachute opens and he descends vertically with uniform retardation for a further 22·5 s. His speed is zero as he reaches the ground.

Calculate the speed of the paratrooper just before his parachute opens.

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Assuming air resistance on the paratrooper without the parachute to be negligible, the speed of the paratrooper just before his parachute opens can be calculated using the following Newton's equation of motion:

`v = u + g*t`

Here, u = 2 m/s, g = 9.81 m/s, t = 1.5 s

Plugging in the values we get:

v = 2 + 9.81 * 1.5 = 16.715 m/s

Therefore, the speed of the paratrooper just before his parachute opens is **16.715 m/s.**