A pole-vaulter converts the kinetic energy of running to elastic potential energ

Question

A pole-vaulter converts the kinetic energy of running to elastic potential energy in the pole, which is then converted to gravitational potential energy. If a pole-vaulter's center of gravity is 0.987 m above the ground while he sprints at 10.0 m/s, what is the maximum height of his center of gravity during the vault? For an extended object, the gravitational potential energy is U = mgh, where h is the height of the center of gravity. (In 1988, Sergei Bubka was the first pole-vaulter ever to clear 6 m.)

Explanation / Answer

Given that

Pole valut center of gravity is cg = 1.3 m

Speed of the pole valuter v = 10.2 m/s

From law of conservation of energy

      KE = PE

        1/2 mv2 = mgh

           (0.5 )v2 = gh

Then the height is

                h = (0.5 )v2 /g

                       = (0.5)(10.2 m/s)2/ 9.8 m/s2

                       = 5.3 m

Since the center of gravity is 1.3 above the ground then we want to add a that because we measurefrom the center of gravity

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