A helicopter is hovering at rest above horizontal ground at the point $$\(H\)$$. A parachutist steps out of the helicopter and immediately falls vertically and freely under gravity from rest for $$\(2.5 \mathrm{~s}\)$$. His parachute then opens and causes him to immediately decelerate at a constant rate of $$\(3.9 \mathrm{~m} \mathrm{~s}^{-2}\)$$ for $$\(T\)$$ seconds $$\((T< 6)\)$$, until his speed is reduced to $$\(V \mathrm{~m} \mathrm{~s}^{-1}\)$$. He then moves with this constant speed $$\(V \mathrm{~m} \mathrm{~s}^{-1}\)$$ until he hits the ground. While he is decelerating, he falls a distance of $$\(73.75 \mathrm{~m}\)$$. The total time between the instant when he leaves $$\(H\)$$ and the instant when he hits the ground is $$\(20 \mathrm{~s}\)$$. The parachutist is modelled as a particle. (a) Find the speed of the parachutist at the instant when his parachute opens. (1) (b) Sketch a speed-time graph for the motion of the parachutist from the instant when he leaves $$\(H\)$$ to the instant when he hits the ground. (2) (c) Find the value of $$\(T\)$$. (5) (d) Find, to the nearest metre, the height of the point $$\(H\)$$ above the ground. (4)

Answer:

Knowledge points:

Solution: