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.
Draw a sketch of the velocity-time graph for the paratrooper’s descent.
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The paratrooper jumps out of the stationary helicopter with an initial velocity of 2 m/s vertically downwards. As he falls for 1.5 seconds, the velocity increases in a linear fashion and reaches 2 + 9.8*1.5 = 16.7 m/s.
After the opening of the parachute, the paratrooper decelerates and his velocity after 22.5 seconds is 0. The resulting deceleration in this period is 167/225 m/s^2 and the decrease is velocity is linear.
The velocity-time graph for the paratrooper's descent is:
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