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

Question

Dr. Abubacker · Accepted Answer

(c) If a stone is dropped from height $h$ then
$h = \frac{1}{2}gt^{2}$ …(i)
if a stone is thrown upward with velocity $u$ then
$h = - u\ t_{1} + \frac{1}{2}gt_{1}^{2}$ …(ii)
If a stone is thrown downward with velocity $u$ then
$h = ut_{2} + \frac{1}{2}gt_{2}^{2}$ …(iii)
From (i) (ii) and (iii) we get
$- ut_{1} + \frac{1}{2}gt_{1}^{2} = \frac{1}{2}gt^{2}$ …(iv)
$ut_{2} + \frac{1}{2}gt_{2}^{2} = \frac{1}{2}gt^{2}$ …(v)
Dividing (iv) and (v) we get
$	herefore\frac{- ut_{1}}{ut_{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$\ t = \sqrt{t_{1}t_{2}}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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