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, ltBrgt `v=u+at" …(1)"`
`and s=ut+(1)/(2)at^(2)" …(2)"`
`therefore s=(v-at)t+(1)/(2)at^(2)`
`=vt-at^(2)+(1)/(2)at^(2)`
`therefore s=vt-(1)/(2)at^(2)" ...(3)"`
As the stone moves upward from `ArarrB, s=AB=h, t=t_(1),`
`a=-g" (retardation),"`
u = u and v = 0
`therefore" From Eq. (3), "h=0-(1)/(2)(-g)t_(1)^(2)`
`therefore h=(1)/(2)g t_(1)^(2)" ...(4)"`
As the stone moves downward from `B rarr A, t=t_(2), u=0, s=h and a=g`
`therefore" form Eq. (2), "h=(1)/(2)g t_(2)^(2)" ....(5)"` ltBrgt From Eqs. (4) and (5), `t_(1)^(2)=t_(2)^(2)`
`therefore t_(1)=t_(2)" "(because t_(1) and t_(2)" are positive")`
we can conclude that the time taken by the stone to go up is same as the time taken to come down .