A ball is dropped from a height. If it takes 0.2s to cross the last 6m before hitting the ground, find the height from which it is dropped. Take `g = 10 m/ /s^(2) (AS_(1))`

Answer

B

Explanation

For last 6 m distance tavelled, `S= 6m, u=?, t=0.2 sec, a=g= 9.8 m/s^2, S= ut+ 1/2 at^2, 6=u(0.2)+4.9xx0.04, rarr u=5.8/0.2, 29 m/s`, For distance `x, u=0, v= 29 m/s, as=g= 9.8 m/s^2, :. S_2 = (v^2-u^2)/(2a), = (29^2-0^2)/(2xx9.8)=(29xx29)/19.6, Total distance = 42.5+6=48.5m = 48`