A balloon starts from rest moves vertically upwards with an acceleration `(g//8) ms^(-2)` a stone falls from the balloon after 8 s from the start. Further time taken by the stone to reach the ground is: `(g=9.8ms^(-2))`

Explanation

The distance of the stone above the ground about which it begins to fall from the balloon is
`h = (1)/(2)((g)/(8))8^(2) = 4g`
The velocity of the balloon at this height can be obtained from `v = u + at, v = 0 + ((g)/(8))8 = g`
This becomes the initial velocity (u) of the stone as the stone falls from the balloon at the height h.
`therefore u. =g`
For the total motion of the stone `h = -u^(1)t + (1)/(2)gt^(2)`
`therefore-4g = gt - (1)/(2)gt^(2) and t^(2) - 2t - 8 = 0`
Solving for .t. we get t = 4 and -2s. Ignoring negative value of time , t = 4s
Three bodies are projected from towers of same height as shown. 1st one is projected vertically up with a velocity .u.. The second one is thrown down vertically with the same velocity and the third one is dropped as a freely falling body. IF `t_(1), t_(2), t_(3)` are the times taken by them to reach ground, then,

For 1st body, `h = -ut_(1) + (1)/(2)gt_(1)^(2) to (1)`
For 2nd body, `h = ut_(2) + (1)/(2)gt_(2)^(2) to (2)`
For 3rd body, `h = (1)/(2)gt_(3)^(2) to (3)`
From `(1) xx t_(2) + (2) xx t_(1)`