A girl in a swing is $2.5 m$ above ground at the maximum height and at $1.5 m$ above the ground at the lowest point. Her maximum velocity in the swing is $(g = 10 {m s}^{- 2})$

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

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

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2 \sqrt{3} {m s}^{- 1}

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

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

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Solution

The correct option is B $2 \sqrt{5} {m s}^{- 1}$
In this case,
Change in potential energy $=$ Kinetic energy
$\frac{1}{2} m v^{2} = m g (h_{2} - h_{1})$
$v^{2} = 2 g (2.5 - 1.5)$
$v^{2} = 2 \times 10 \times 1$
$v = \sqrt{5 \times 2 \times 2}$
$v = 2 \sqrt{5} m / s$

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A girl in a swing is 2.5m above ground at the maximum height and at 1.5m above the ground at the lowest point. Her maximum velocity in the swing is (g=10ms−2)

The correct option is B 2√5ms−1In this case,Change in potential energy = Kinetic energy12mv2=mg(h2−h1)v2=2g(2.5−1.5)v2=2×10×1v=√5×2×2v=2√5m/s

A girl in a swing is $2.5 m$ above ground at the maximum height and at $1.5 m$ above the ground at the lowest point. Her maximum velocity in the swing is $(g = 10 {m s}^{- 2})$

The correct option is B $2 \sqrt{5} {m s}^{- 1}$
In this case,
Change in potential energy $=$ Kinetic energy
$\frac{1}{2} m v^{2} = m g (h_{2} - h_{1})$
$v^{2} = 2 g (2.5 - 1.5)$
$v^{2} = 2 \times 10 \times 1$
$v = \sqrt{5 \times 2 \times 2}$
$v = 2 \sqrt{5} m / s$