A boy on a cliff $500 m$ high drops a stone. Three seconds later he throws another stone. They both hit the ground at the same time. The speed with which he threw the second stone is nearly equal to:
(take $g = 10 m / s^{2}$ ) (Neglect air friction)

0 m / s

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36.43 m / s

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106.43 m / s

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32.67 m / s

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Solution

The correct option is B $36.43 m / s$
If vertically upwards direction is taken as positive axis, then
$h = - 500 m, a = - g = - 10 m / s^{2}$ ,
$u = 0, h = u t + \frac{1}{2} a t^{2}$

$- 500 = 0 - 5 \times t^{2} \Rightarrow t = 10 s$

The second stone is thrown 3 seconds later, hence time taken for second stone to reach the ground is $10 - 9 = 7 s$ .

$h = u t + \frac{1}{2} g t^{2} \Rightarrow - 500 = - u \times 7 - 5 \times 7^{2} 7 u = 500 - 5 \times 49 \Rightarrow u = 36.43 m / s$

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A boy on a cliff 500 m high drops a stone. Three seconds later he throws another stone. They both hit the ground at the same time. The speed with which he threw the second stone is nearly equal to: (take g=10 m/s2) (Neglect air friction)

A boy on a cliff $500 m$ high drops a stone. Three seconds later he throws another stone. They both hit the ground at the same time. The speed with which he threw the second stone is nearly equal to:
(take $g = 10 m / s^{2}$ ) (Neglect air friction)