Question

A 52-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.2 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.

Answer 1

The altitude is 5.03 m

Step-by-step explanation:

Given:

Mass of pole
`m = 52` kg

Speed of vaulter
`v_(i) = 10(m)/(s)`

Speed of vaulter when she is above the bar
`v_(f) = 1.2 (m)/(s)`

For finding altitude we use conservation of energy,

`\Delta E _(i) = \Delta E _(f)`

`(1)/(2) m v_(f) ^(2) + mgh= (1)/(2) m v _(i) ^(2) +0` ( ∵ initial height is zero )

`v_(i) ^(2) - v_(f) ^(2) = 2g h`

`h = (v_(i) ^(2) - v_(f) ^(2))/(2g)`

`h = (100-1.44)/(2 * 9.8)` ( ∵
`g = 9.8 (m)/(s^(2) )` )

`h = 5.03` m

Therefore, the altitude is 5.03 m