A boy standing at the top of a tower of 20 $m$ height drops a stone. Assuming $g = 10 m s^{- 2}$ , the velocity with which it hits the ground is:

20 m s^{- 1}

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40 m s^{- 1}

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5 m s^{- 1}

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10 m s^{- 1}

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Solution

The correct option is A $20 m s^{- 1}$
When there is a free fall, we can directly use the equation of motion :
$v^{2} = u^{2} + 2 g h$
here $u = 0$
so, $v = \sqrt{2 g h}$ = $\sqrt{2 \times 10 \times 20} = 20 m / s$

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A boy standing at the top of a tower of 20 m height drops a stone. Assuming g=10ms−2, the velocity with which it hits the ground is:

The correct option is A 20ms−1When there is a free fall, we can directly use the equation of motion :v2=u2+2ghhere u=0so, v=√2gh = √2×10×20=20 m/s

A boy standing at the top of a tower of 20 $m$ height drops a stone. Assuming $g = 10 m s^{- 2}$ , the velocity with which it hits the ground is:

The correct option is A $20 m s^{- 1}$
When there is a free fall, we can directly use the equation of motion :
$v^{2} = u^{2} + 2 g h$
here $u = 0$
so, $v = \sqrt{2 g h}$ = $\sqrt{2 \times 10 \times 20} = 20 m / s$