A stone dropped from a building of height h and it reaches after … - Vidyadip

MCQ

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

A
$t = {t_1} - {t_2}$
B
$t = \frac{{{t_1} + {t_2}}}{2}$
✓
$t = \sqrt {{t_1}{t_2}} $
D
$t = t_1^2t_2^2$

✓

Answer

Correct option: C.

$t = \sqrt {{t_1}{t_2}} $

c
(c) 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}} $

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