Question

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

Answer 1

We have v = u + at

and s = ut + `1/2 "at"^2`

∴ s = `("v" - "at")"t" + 1/2"at"^2 + 1/2"at"^2`

s = ut - `1/2 "at"^2`

As the stone moves upward from A → B

S = AB = h, t = t₁, a = - g (retardation)

u = u and v = 0

h = 0 - `1/2 (- "g")"t"_1^2`

h = `1/2"gt"_1^2`

As the stone moves downward from  B → A

t = t₂, u = 0, s = h and a = g

h = `1/2"gt"_2^2`

`"t"_1^2 = "t"_2^2`

`"t"_1 = "t"_2`