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A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest point she is 0.8 m from the ground. What is her maximum speed?The acceleration of gravity is 9.8 m/s². Answer in units of m/s.

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usmanbiu

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + [tex]0.5mV1^{2}[/tex] = mgh2 + [tex]0.5mV2^{2}[/tex]

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + [tex]0.5mV1^{2}[/tex] = mgh2

      m(gh1 + [tex]0.5V1^{2}[/tex]) = m(gh2)

      gh1 + [tex]0.5V1^{2}[/tex] = gh2

      V1 = [tex]\sqrt{\frac{gh2 - gh1}{0.5}}[/tex]

now we can substitute all required values into the equation above.

V1 = [tex]\sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}[/tex]

V1 = [tex]\sqrt{\frac{32.34}{0.5}}[/tex]

V1 =8.1 m/s

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