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Chapter 6: Problem 78

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 \(1.0 \mathrm{~m}\) above the ground while he sprints at \(10.0 \mathrm{~m} / \mathrm{s},\) what is the maximum height of his center of gravity during the vault? For an extended object, the gravitational potential energy is \(U=m g h,\) where \(h\) is the height of the center of gravity. (In \(1988,\) Sergei Bubka was the first pole-vaulter ever to clear \(6 \mathrm{~m}\).)

Short Answer

Expert verified
Answer: The maximum height of the pole-vaulter's center of gravity during the vault is approximately 6.1 meters.

Step by step solution

01

Determine the initial kinetic energy of the pole-vaulter

To find the initial kinetic energy, we'll use the formula \(KE = \frac{1}{2} mv^2\) where m is the mass of the pole-vaulter and v is his speed. Since we are not given the pole-vaulter's mass, we can't find an exact value for the kinetic energy at this point. However, we can express it in terms of m: \(KE = \frac{1}{2} m(10.0 \mathrm{~m/s})^2\)
02

Set up the conservation of energy equation

The conservation of energy equation will involve the initial kinetic energy, the elastic potential energy in the pole, and the final gravitational potential energy at the highest point of the vault. The equation will look like this: \(KE + PE_{elastic} = PE_{gravitational}\) Since we're not given any information about the pole or its elasticity, we can assume that all of the initial kinetic energy is converted into gravitational potential energy: \(\frac{1}{2} m(10.0 \mathrm{~m/s})^2 = mgh\)
03

Solve for the maximum height h

To find the maximum height of the vaulter's center of gravity during the jump, we'll solve for h in the conservation of energy equation: \(\frac{1}{2} m(10.0 \mathrm{~m/s})^2 = mgh\) Divide both sides by m to cancel it out: \(\frac{1}{2}(10.0 \mathrm{~m/s})^2 = gh\) Now divide by g (approx. \(9.81 \mathrm{~m/s^2}\)) to solve for h: \(h = \frac{1}{2} \frac{(10.0 \mathrm{~m/s})^2}{9.81 \mathrm{~m/s^2}}\) \(h \approx 5.1 \mathrm{m}\)
04

Calculate the total height of the pole-vaulter's center of gravity

Now that we have found the maximum height h reached during the vault, we also need to add the initial height of the center of gravity (1.0 m) to get the total height: \(total\_height = h + initial\_height\) \(total\_height = 5.1 \mathrm{m} + 1.0 \mathrm{m}\) \(total\_height \approx 6.1 \mathrm{m}\) Therefore, the maximum height of the pole-vaulter's center of gravity during the vault is approximately 6.1 meters.

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