A ball rolled off the edge of a horizontal table with a speed of $4 m s^{- 1}$ .if it reaches the ground after $0.3$ sec. Then the speed with which it strikes the ground is $(g = 10 m s^{- 2}$ )

4 \sqrt{5} m s^{- 1}

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5 m s^{- 1}

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5 \sqrt{2} m s^{- 1}

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3 m s^{- 1}

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Solution

The correct option is B $5 m s^{- 1}$
$v^{2} = u^{2} + 2 a 5$
$v^{2} = y^{2} + 2 \times 10 \times 0.45$
$= 16 + 9$
$v^{2} = 25$
$v = 5$

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