A stone dropped from a building of height $h$ and it reaches after $t$ seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity $u$ and they reach the earth surface after ${t_1}$ and ${t_2}$ seconds respectively, then
$t = {t_1} - {t_2}$
$t = \frac{{{t_1} + {t_2}}}{2}$
$t = \sqrt {{t_1}{t_2}} $
$t = t_1^2t_2^2$
A stone dropped from a building of height $h$ and it reaches after $t$ seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity $u$ and they reach the earth surface after ${t_1}$ and ${t_2}$ seconds respectively, then
If a stone is dropped from height $h$
then $h = \frac{1}{2}g\,{t^2}$…(i)
If a stone is thrown upward with velocity $u$ then
$h = - u\;{t_1} + \frac{1}{2}g\;t_1^2$…(ii)
If a stone is thrown downward with velocity $u$ then
$h = u{t_2} + \frac{1}{2}gt_2^2$…(iii)
From (i) (ii) and (iii) we get
$ - u{t_1} + \frac{1}{2}g\,t_1^2 = \frac{1}{2}g\,{t^2}$…(iv)
$u{t_2} + \frac{1}{2}g\,t_2^2 = \frac{1}{2}g\,{t^2}$…(v)
Dividing (iv) and (v) we get
$\therefore \frac{{ - u{t_1}}}{{u{t_2}}} = \frac{{\frac{1}{2}g({t^2} - t_1^2)}}{{\frac{1}{2}g({t^2} - t_2^2)}}$
or $ - \frac{{{t_1}}}{{{t_2}}} = \frac{{{t^2} - t_1^2}}{{{t^2} - t_2^2}}$
By solving $t = \sqrt {{t_1}{t_2}} $
then $h = \frac{1}{2}g\,{t^2}$…(i)
If a stone is thrown upward with velocity $u$ then
$h = - u\;{t_1} + \frac{1}{2}g\;t_1^2$…(ii)
If a stone is thrown downward with velocity $u$ then
$h = u{t_2} + \frac{1}{2}gt_2^2$…(iii)
From (i) (ii) and (iii) we get
$ - u{t_1} + \frac{1}{2}g\,t_1^2 = \frac{1}{2}g\,{t^2}$…(iv)
$u{t_2} + \frac{1}{2}g\,t_2^2 = \frac{1}{2}g\,{t^2}$…(v)
Dividing (iv) and (v) we get
$\therefore \frac{{ - u{t_1}}}{{u{t_2}}} = \frac{{\frac{1}{2}g({t^2} - t_1^2)}}{{\frac{1}{2}g({t^2} - t_2^2)}}$
or $ - \frac{{{t_1}}}{{{t_2}}} = \frac{{{t^2} - t_1^2}}{{{t^2} - t_2^2}}$
By solving $t = \sqrt {{t_1}{t_2}} $
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