A parachutist, after bailing out, falls `50 m` without friction, When the parachute opens, it decelerates at `2 ms^(-2)`. He reaches the ground with a speed of `3 m s^(-1)`. At what height did he bail out?

Options

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

Answer

B

Explanation

After bailing out from position A, parachutist falls freely under gravity a distance 50m. Let the velocity acquired by him at position B be ‘v’.
` v^(2) =u^(2) +2as =0 +2xx 9.8 xx 50 =980`
[As `u=0 , a=9.8 m//s ^(2) ,s=50m ]` At position B, parachute opens and it moves with retardation of 2`m//s^(2)` and reach at ground (position C) with velocity of 3m/s For the journey of the part ‘BC’ again applying the equation ` v^(2) =u^(2) +2as , `where
` v= 3 m //s ,u =sqrt( 980 )m //s ,a =- 2m //s ^(2) s=h`
`rArr " "(3)^(2) =(sqrt(980)) ^(2) +2xx (-2) xx h rArr 9= 980 -4h rArr " "h= ( 980-9)/( 4) =( 971)/(4) =242.7 = 24 m ` So, the total height by which parachutist bailed out = 50 + 243 = 293 m.