A stone thrown vertically upwards with initial velocity u reaches a height h before coming down. Show that the time taken by it to go up is the same as the time taken to come down.

Explanation

We have ` v = u + at`
and ` s = ut + 1/2 at^(2)`
` therefore s = ( v-at) t + 1/2 at^(2) = vt-at^(2) + 1/2 at^(2)`
` s = ut - 1/2 at^(2)`
As the stone moves upward from ` A to B`
S = AB = h, t = ` t_(1)`, a= - g ( retardation )
u = u and v = 0
From Eq . (3), h = ` 0 - 1/2 (-g) t_(1)^(2)`
` h = 1/2 "gt"_(1)^(2)`
As the stone moves downward from ` B to A `
` t = t_(2) , u = 0 , s = h and a =g `
From Eq. (2). ` h = 1/2 "gt"_(2)^(2)`
From Eqs. (4) and (5) , `t_(1)^(2) = t_(2)^(2)`
` t_(1) = t_(2) `