Question

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.

Answer 1

V1 =8.1 m/s

Step-by-step 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 +
`0.5mV1^(2)` = mgh2 +
`0.5mV2^(2)`

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 +
  `0.5mV1^(2)` = mgh2

m(gh1 +
`0.5V1^(2)`) = m(gh2)

gh1 +
`0.5V1^(2)` = gh2

V1 =
`\sqrt{(gh2 - gh1)/(0.5)}`

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

V1 =
`\sqrt{((9.8x4.1) - (9.8x0.8))/(0.5)}`

V1 =
`\sqrt{(32.34)/(0.5)}`

V1 =8.1 m/s