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August 29, 2014

August 29, 2014

Posted by **Chad** on Wednesday, August 8, 2012 at 10:09pm.

2 in. (She did less than that to win gold this summer.) If all of her gravitational potential energy came only

from her kinetic energy, calculate (in m/s) the running speed necessary to make this vault. Assuming the

vaulting pole to be elastic, does the pole “add”energy to this system? Explain.

- Physics -
**drwls**, Wednesday, August 8, 2012 at 10:53pmH = 16 ft, 2 inches = 194 in. = 4.9276 m

Set M g H = (1/2) M Vo^2 and solve for the required initial velocity Vo. Note that the mass cancels out.

Vo = sqrt(2 g H) = 9.83 m/s

This formula assumes that the center of mass of the vaulter is raised the full amount H. Actually, because of the way the body bends over the bar, the vaulter's center of mass passes under the bar if the best vaulting technique is used. Also, the vaulter starts the vault with his of her center of mass already about 0.9 meters above the ground. In a 4.93 meter vault, the center of mass only gets increased by about 3.9 meters. This reduces Vo to about 8.7 m/s, more typical of an actual vault by a world class vaulter.

With respect to the question about the flexible elastic pole, this does not add energy to the system, but it allows most of the kinetic energy to be stored as pole elastic deformation energy and later released at the maximum height of the vault.

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