A balloon starts rising from the ground with an acceleration of `1.25ms^-2`. After 8 seconds, a stone is released from the balloon. After releasing, the stone will:

Options

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

Answer

C

Explanation

at t=8sec
height of balloon
`h=0+(1)/(2)xx1.25xx(8)^(2)=40m`
and velocity of balloon
`v_(b)=0+1.25xx8=10 m//s` (upward)
and as stone is released, it has the same initial velocity i.e., 10 m/sec (upward) hence it will go up by `5m[h=(u^(2))/(2g)=(10xx10)/(2xx10)=5m]` and come back 5m and then fall down a distance 40m downwards So total distance covered by stone
`=40+5+5=50m`
and displacement of stone is 40m downwards.
Now if time of journey is t then
`h=yt+(1)/(2)at^(2)rArr40=-10t+(1)/(2)xx10xxt^(2)`
`rArr8=-2t+t^(2)rArrt^(2)-2t-8=0`
`(t-4)(t+2)=0rArrt=4`sec(t=-22 not possible)