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
We have, $v = u + at …..(1)$
and $s = ut + \frac{1}{2} at^2 …..(2)$
$\therefore s = (v –at) t + \frac{1}{2} at^2$
$= vt –at^2 + \frac{1}{2} at^2$
$\therefore s = vt –\frac{1}{2} at^2 …..(3)$
As the stone moves upward from $A \rightarrow B,$
$s = AB = h, t = t_1 ,$
$a = -g ($retardation$),$
$u = u$ and $v = 0$
$\therefore $ From Eq. $(3), h = 0 –\frac{1}{2} (-g)t_1 ^2$
$\therefore h = \frac{1}{2}gt_1 ^2 …..(4)$
As the stone moves downward from $B \rightarrow A,$
$t = t_2 , u = 0, s = h$ and $a = g$
$\therefore $ from Eq. $(2), h = \frac{1}{2} gt^2 …..(5)$
From Eqs. $(4)$ and $(5), t_1 ^2 = t_2^2$
$\therefore t_1 = t_2 (\because t_1 $ and $t_2$_ are positive$)$
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