Question

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

Answer 1

A → B, the stone moves upward; B → A; the stone moves downward

We have, v = u + at …..(1)

and s = ut + \(\frac{1}{2}\) at² …..(2)

∴ s = (v – at) t + \(\frac{1}{2}\) at²

∴ s = vt – \(\frac{1}{2}\) at² …..(3)

As the stone moves upward from A → B,

s = AB = h, t = t₁,

a = -g (retardation),

u = u and v = 0

∴ From Eq. (3), h = 0 – \(\frac{1}{2}\) (-g)t₁²

∴ h = \(\frac{1}{2}\) gt₁² …..(4)

As the stone moves downward from B → A,

t = t₂, u = 0, s = h and a = g

∴ from Eq. (2), h = \(\frac{1}{2}\) gt² …..(5)

From Eqs. (4) and (5), t₁² = t₂²

∴ t₁ = t₂ (∵ t₁ and t₂ are positive)