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

Options

A. 293 m
B. 111 m
C. 91 m
D. 182 m

Answer

A

Explanation

After bailing out from point `A` parachutist fails freely under gravity. The velocity acquired by it will 'v'
From `v^2 = u^2 + 2as = 0 + 2 xx 9.8 xx 50 = 980`
`["As" u = 0, a = 9.8 m//s^2, s = 50 m]`
At point `B`, parachute opens and it moves with retardation of `2 m//s^2` and reach at ground (Point C) with velocity of `3 m//s`
For the part 'BC' by applying the equation
`v^2 = u^2 + 2as`
`v = 3 m//s, u = sqrt(980) m//s, a = -2 m//s^2, s = h`
`rArr (3)^2 = (sqrt(980))^2 + 2 xx (-2) xx h rArr 9 = 980 - 4h`
`rArr h = (980 - 9)/(4) = (971)/(4) = 242.7 = 243 m`
So, the total height by whicg parachutist bail out `= 50 + 243 = 293 m`.