A Child Is Swinging A Swing Minimum And The Maximumheights Of Swing From Earth' S Surface Are 0.75 M And 2 M Respectively. What Will Be The Maximum Velocity Of This Swing?


From energy conservation \( \frac{1}{2}mv^{2} max = mg (H_{2}H_{1})\)


H1 = minimum height of swing from earth’s surface= 0.75 m

H2 = maximum height of swing from earth’s surface = 2 m

Therefore, \( \frac{1}{2} mv^{2}_{max} = mg (2 – 0.75)\\v_{max} = \sqrt{2 * 10 * 1.25}\\ v_{max} = \sqrt{25}\\ v_{max} = 5ms^{-1}\)

Therefore, the maximum velocity of this swing will be 5ms-1

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