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Question: A stone dropped from a building of height h and it reaches after t seconds on earth. From the same b...

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 t1t _ { 1 } and t2t _ { 2 } seconds respectively, then

A

t=t1−t2t = t _ { 1 } - t _ { 2 }

B

t=t1+t22t = \frac { t _ { 1 } + t _ { 2 } } { 2 }

C

t=t1t2t = \sqrt { t _ { 1 } t _ { 2 } }

D

t=t12t22t = t _ { 1 } ^ { 2 } t _ { 2 } ^ { 2 }

Answer

t=t1t2t = \sqrt { t _ { 1 } t _ { 2 } }

Explanation

Solution

For first case of dropping h=12gt2h = \frac { 1 } { 2 } g t ^ { 2 }

For second case of downward throwing

h=−ut1+12gt12\mathrm { h } = - \mathrm { ut } _ { 1 } + \frac { 1 } { 2 } \mathrm { gt } _ { 1 } ^ { 2 } =12gt2= \frac { 1 } { 2 } g t ^ { 2 }

⇒ −ut1=12g(t2−t12)- u t _ { 1 } = \frac { 1 } { 2 } g \left( t ^ { 2 } - t _ { 1 } ^ { 2 } \right) ......(i)

For third case of upward throwing h=ut2+12gt22=12gt2h = u t _ { 2 } + \frac { 1 } { 2 } g t _ { 2 } ^ { 2 } = \frac { 1 } { 2 } g t ^ { 2 }

⇒ ut2=12g(t2−t22)u t _ { 2 } = \frac { 1 } { 2 } g \left( t ^ { 2 } - t _ { 2 } ^ { 2 } \right) .......(ii)

on solving these two equations : −t1t2=t2−t12t2−t22- \frac { t _ { 1 } } { t _ { 2 } } = \frac { t ^ { 2 } - t _ { 1 } ^ { 2 } } { t ^ { 2 } - t _ { 2 } ^ { 2 } }

⇒ t=t1t2t = \sqrt { t _ { 1 } t _ { 2 } }