A parachutist after bailing out falls $$50$$ m without friction when parachute opens, it decelerates at $$2$$ $$ms-2$$. He reaches the ground with speed of $$3$$ $$m/$$ s. At what height did he bail out?

Question

A parachutist after bailing out falls $$50$$ m without friction when parachute opens, it decelerates at $$2$$ $$ms-2$$. He reaches the ground with speed of $$3$$ $$m/$$ s. At what height did he bail out?

Infinity Learn · Accepted Answer

After bailing out from point A, parachutist falls freely under gravity. The velocity acquired by it will be v.From $$v2$$ $$=$$ $$u2$$ $$+$$ $$2as$$ $$=$$ $$0$$ $$+$$ $$2$$ x $$9.8$$ x $$50$$ $$=$$ $$980$$                                        (As$$, u $$=$$ $$0$$, a $$=$$ $$9.8$$ $$ms-2$$ , s $$=$$ $$50$$ $$m)At$$ point B, parachute opens and it moves with retardation of $$2$$ $$ms-2$$ and reach at ground $$(point$$ C with velocity of $$3$$ $$ms-1$$ $$).For the part BC by applying the equation $$v2$$ $$=$$ $$u2$$ $$+$$ $$2as$$                $$v=3$$ $$ms-1$$,$$u=980$$ $$ms-1$$,$$a=-2$$ $$ms-2$$,$$s=h$$⇒      (3)2=(980)2+2$$×$$(-2)×h⇒         $$9=980-4h$$⇒        $$h=980-94=9714=242.7$$≅$$243$$ mSo, the total height by which parachutist bails out              $$=$$ $$50$$ $$+$$ $$243$$ $$=$$ $$293$$ m

A parachutist after bailing out falls 50 m without friction when parachute opens, it decelerates at 2 ms-2. He reaches the ground with speed of 3 m/ s. At what height did he bail out?

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