A balloon starts rising from the ground with a constant acceleration of `1.25m//s^(2)`. After 8 s, a stone is released from the balloon. Find the time taken by the stone to reach the ground. (Take `g=10m//s^(2)`)

Options

A. The initial velocity of the particle is u
B. The acceleration of the particle is a
C. The acceleration of the particle is 2a
D. At `t=2s` particle is at the origin

Answer

C, D

Explanation

`x=u(t-2)+a(t-2)^(2)`…(i)
`v=(dx)/(dt)=u+2a(t-2)`
Therefore `v(0)=u-4a`
`a=(d^(2)x)/(dt^(2))=2a" "` Hence [C]
`x(2)=0`[From (i)]. Hence [D]