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 height of the paratrooper above the ground when he jumped out of thehelicopter.
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The initial velocity of the paratrooper when he jumped out of the stationary helicopter was 2 m/s vertically downwards. He fell freely under gravity for 1.5 s, then his parachute opened and he descended vertically with uniform retardation for a further 22.5 s. His final velocity when he reached the ground was zero.
The distance traveled by the paratrooper in the first 1.5 s after he jumped down can be calculated using the formula s = u*t + (1/2)*a*t^2, Substituting the values given, s = 2*1.5 + (1/2)*9.8*1.5^2 = 14.025.
The velocity of the pilot when the parachute opened was 2 + 1.5*9.8 = 16.7 m/s in a direction vertically downwards. For the next 22.5 s his velocity is decreased and brought down to 0. The resulting acceleration is 16.7/22.5 m/s^2 in a direction vertically upwards. The distance traveled by the paratrooper during this period of time is 16.7*22.5 - (1/2)*(167/225)*22.5^2 = 187.875
This gives the height of the paratrooper when he jumped off the helicopter as 201.9 m
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