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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.
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Solution
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.)
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