Question

A 70-kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.3 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

`h_(B) = 6.083\,m`

Step-by-step explanation:

Let assume that pole vaulter begins running at a height of zero. The pole vaulter is modelled after the Principle of Energy Conservation:

`K_(A) = K_(B) + U_(g,B)`

`(1)/(2)\cdot m \cdot v_(A)^(2) = (1)/(2)\cdot m \cdot v_(B)^(2) + m\cdot g \cdot h_(B)`

The expression is simplified and final height is cleared within the equation:

`(1)/(2)\cdot (v_(A)^(2) - v_(B)^(2)) = g\cdot h_(B)`

`h_(B) = ((v_(A)^(2)-v_(B)^(2)))/(2\cdot g)`

`h_(B) = ([(11\,(m)/(s) )^(2)-(1.3\,(m)/(s) )^(2)])/(2\cdot (9.807\,(m)/(s^(2)) ))`

`h_(B) = 6.083\,m`