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. (a) ` 11 m`
B. (b) ` 29 3 m`
C. (c ) ` 1 82 m`
D. (d)` 243 m`

Answer

B

Explanation

First `50` m fall is under the effect of gravity only. The velocity acquired
` u= sqrt 2 gh ) = sqrt ( 2 xx 9.8 xx 500 = 10 sqrt (9.8) ms^(-1)`
Taking onward motion of prachutist with retardation ` 2 m//s^2` , v = 3 have , u = 10 sqrt 9.8 m//s ,
` a =- 2 ms^(-1) , we have , ` u = 10 sqrt 9.8 m//s ,
` a=- 2 ms^(-1) , v= 3 m//s , S= ?`
S= (v^2 -u^2)/(2 a) = (3^2 - (100 xx 9.8)) /( 2 xx (-2) = 243 m`.
Total height ` = 50 + 243 = 293 m`.