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                 ...(1)

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

∴ s = `("v" - "at")"t" + 1/2"at"^2`

= `"vt" - "at"^2+1/2"at"^2`

∴ `"s" = "vt" - 1/2 "at"^2`                       ...(3)

As the stone moves upward from (A → B) [AB = h]

S = AB = h, t = t₁,

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`                               ...(4)

As the stone moves downward from (B → A)

t = t₂, u = 0, s = h and a = g

∴ From Eq. (2), h = `1/2"gt"_2^2`           ...(5)

From Eqs. (4) and (5), 

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

∴ `"t"_1 = "t"_2`   (∵ t₁, t₂ are positive.)