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

Answer:

$h_{B} = 6.083\,m$

Explanation:

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

$K_{A} = K_{B} + U_{g,B}$

$\frac{1}{2}\cdot m \cdot v_{A}^{2} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + m\cdot g \cdot h_{B}$

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

$\frac{1}{2}\cdot (v_{A}^{2} - v_{B}^{2}) = g\cdot h_{B}$

$h_{B} = \frac{(v_{A}^{2}-v_{B}^{2})}{2\cdot g}$

$h_{B} = \frac{[(11\,\frac{m}{s} )^{2}-(1.3\,\frac{m}{s} )^{2}]}{2\cdot (9.807\,\frac{m}{s^{2}} )}$

$h_{B} = 6.083\,m$