Assume that a pole vaulter can convert all of his kinetic energy into potential energy. If a 70.0 kg pole vaulter approaches the vault with a velocity of 9.80 m/s, about how high can he get?

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Correct Answer:

**B.** 4.90 m.

Conservation of energy means the sum of the kinetic and potential energies must be the same at the start of the jump and at its maximum height:

1/2 mv^{2} + mgh = 1/2 mv^{2} + mgh
(Initial energies on the left and final energies on the right.)

There is no height initially, so there is no initial potential energy. At the top of the jump there is no velocity, so there is no final kinetic energy. This gives:

1/2 mv^{2} + 0 = 0 + mgh
Therefore:

h = v^{2} / 2g = (9.8 m/s)^{2} / 2(9.8 m/s^{2}) = 4.9 m
The mass of the vaulter cancels out.

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Photo attributed to Marquis Lewis.

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