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  • A stone dropped from a building of height reaches the ground after <img alt="t" src="https://entrancecorner.oncodecogs.com/gi

A stone dropped from a building of height h reaches the ground after t seconds. From the same building if two stones are thrown (one upwards and the other downwards) with the same velocity u and they reach the ground after t_1 and t_2 seconds respectively, then the time interval t is

Option: 1

\quad t=t_1-t_2


Option: 2

\quad t=\frac{t_1+t_2}{2}


Option: 3

t=\sqrt{t_1 t_2}


Option: 4

t=\sqrt{t_1^2-t_2^2}


Answers (1)

best_answer

For the stone dropped with zero initial velocity, we have

h=\frac{1}{2} g t^2

For the stone thrown upwards with velocity u, we have

h=-u t_1+\frac{1}{2} g t_1^2

For the stone thrown downwards with velocity u, we have

h=u t_2+\frac{1}{2} g t^2

Notice that the displacement of the stone in the three case is the same, equal to h. We have

\frac{1}{2} g t^2=-u t_1+\frac{1}{2} g t_1^2 \Rightarrow u t_1=\frac{1}{2} g t_1^2-\frac{1}{2} g t^2

and  \frac{1}{2} g t^2=u t_2+\frac{1}{2} g t_2^2 \Rightarrow u t_2=\frac{1}{2} g t^2-\frac{1}{2} g t_2^2 Dividing the two equations, we have

\frac{t_1}{t_2}=\frac{t_1^2-t^2}{t^2-t_2^2}

which gives t=\sqrt{t_1 t_2} .

Hence the correct choice is (c).

Posted by

Kuldeep Maurya

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