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

20 {ms}^{- 1}

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40 {ms}^{- 1}

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5 {ms}^{- 1}

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10 {ms}^{- 1}

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Solution

The correct option is A $20 {ms}^{- 1}$
Given, $g = 10 {ms}^{- 2}$ and $h = 20 m$
Using third equation of motion, we have
$v^{2} - u^{2} = 2 g h$
$\Rightarrow v = \sqrt{2 g h} (∵ u = 0)$
$= \sqrt{2 \times 10 \times 20}$
$= \sqrt{400} = 20 {ms}^{- 1}$

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

The correct option is A 20 ms−1Given, g=10 ms−2 and h=20 m Using third equation of motion, we have v2−u2=2gh ⇒v=√2gh (∵u=0) =√2×10×20 =√400=20 ms−1

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

The correct option is A $20 {ms}^{- 1}$
Given, $g = 10 {ms}^{- 2}$ and $h = 20 m$
Using third equation of motion, we have
$v^{2} - u^{2} = 2 g h$
$\Rightarrow v = \sqrt{2 g h} (∵ u = 0)$
$= \sqrt{2 \times 10 \times 20}$
$= \sqrt{400} = 20 {ms}^{- 1}$