A partachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at `2 ms^(-2)`. At what height did he bail out

Options

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

Answer

A

Explanation

After bailing out from point A partachutist falls freely under gravity. The velcity acquired by it will be `upsilon`.
From `upsilon^(2)+2as=0+2xx9.8xx50=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)` are reach at ground (Point C with velocity of `3 ms^(-1)`). For the part BC by applying the equastion `upsilon^(2)=u^(2)+2as`
`upsilon=3 ms^(-1), u = sqrt(989)ms^(-1), a=-2 ms^(-2), s=h`
`rArr (3)^(2)=(sqrt(980))^(2)+2xx(-2)xxh`
`rArr 9=980-4h`
`rArr h=(980-9)/(4)=(971)/(4)=242.7 ~== 243m`
So, the total height by which parachutist bails out
`=50+243=293m`