Question

During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What minimum speed did he need at launch if he was traveling at 6.7 m/s at the top of the arc

Answer 1

Answer:

1) $v_{A} \approx 8.272\,\frac{m}{s}$

Explanation:

1) Let assume that the campion begins running at a height of zero. The movement of the Olympic champion is modelled after the Principle of Energy Conservation:

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

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

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

The minimum speed is obtained herein:

$v_{A}=\sqrt{v_{B}^{2} + 2 \cdot g \cdot h}$

$v_{A} = \sqrt{(6.7\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)}$

$v_{A} \approx 8.272\,\frac{m}{s}$