A boy is sitting on a swing at a maximum height of 5m above the ground. When the swing passes through the mean position which is 2m above the ground his velocity is approximately

7.6 m/s

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9.8 m/s

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6.26 m/s

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None of these

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Solution

The correct option is A
7.6 m/s

By the conservation of energy Total energy at point A = Total energy at point B

$\Rightarrow m g h_{1} = m g h_{2} + \frac{1}{2} m v^{2}$

$\Rightarrow 9.8 \times 5 = 9.8 \times 2 + \frac{1}{2} v^{2}$

$\Rightarrow v^{2} = 58.8 ∴ v = 7.7 m / s (a p p x .)$

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A boy is sitting on a swing at a maximum height of 5m above the ground. When the swing passes through the mean position which is 2m above the ground his velocity is approximately

The correct option is A 7.6 m/s By the conservation of energy Total energy at point A = Total energy at point B ⇒mgh1=mgh2+12mv2 ⇒9.8×5=9.8×2+12v2 ⇒v2=58.8 ∴v=7.7m/s(appx.)

The correct option is A
7.6 m/s

By the conservation of energy Total energy at point A = Total energy at point B

$\Rightarrow m g h_{1} = m g h_{2} + \frac{1}{2} m v^{2}$

$\Rightarrow 9.8 \times 5 = 9.8 \times 2 + \frac{1}{2} v^{2}$

$\Rightarrow v^{2} = 58.8 ∴ v = 7.7 m / s (a p p x .)$