A balloon starts rising from thee ground with an acceleration of `1.25 m//s^(2)`. After 8 s, a stone is released from the balloon. The stone will (taking `g = 10 ms^(-2)`)

Options

A. have a displacement of 50 m
B. cover a distance of 40 m in reaching the ground
C. reach the ground in 4 s
D. begin to move down after released.

Answer

B, C

Explanation

Velocity of the balloon after 8s `=1.25 xx 8 = 10 m//s`
Height reached `=(v^(2))/(2a) = (100)/(2 xx 1.25) = 40m`
Now the stone will reach the ground in? S= 40m, u= `-10 m//s, a = 10 ms^(-2), t =? 40 = -10t + 1//2 xx 10 xx t^(2)`
Solving we get, t=4s