A boy on a 784 m high cliff drops a stone. One second later, he throws another stone downwards with some speed. The two stones reach the ground simultaneously. Find the speed with which the second stone was thrown.

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Solution

For first stone-

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

$⟹ t = \sqrt{\frac{2 h}{g}}$

$⟹ t_{1} = \sqrt{\frac{2 \times 784}{9.8}} = 12.65 s$

For $2^{n d}$ stone-

$t_{2} = 11.65 s$

distance is same-

$u t_{2} + \frac{1}{2} g t_{2}^{2} = \frac{1}{2} g t_{1}^{2}$

$⟹ u t_{2} = \frac{1}{2} g (t_{1} + t_{2}) (t_{1} - t_{2})$

$⟹ u = \frac{g}{2 t_{2}} (t_{1} + t_{2})$

$⟹ u = 10.22 m / s$

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A boy on a 784 m high cliff drops a stone. One second later, he throws another stone downwards with some speed. The two stones reach the ground simultaneously. Find the speed with which the second stone was thrown.

For first stone-h=12gt2⟹t=√2hg⟹t1=√2×7849.8=12.65sFor 2nd stone-t2=11.65sdistance is same-ut2+12gt22=12gt21⟹ut2=12g(t1+t2)(t1−t2)⟹u=g2t2(t1+t2)⟹u=10.22m/s