MCAT Question A Day - 3/12/14 - Answer!
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 vault?
A. 2.45 m
B. 4.90 m
C. 9.80 m
D. 19.6 m
h = v02 / 2g = (9.8 m/s)2 / 2(9.8 m/s2) = 4.9 m
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The correct answer is (B). 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 mv02 + mgh0 = 1/2 mv2 + mgh
where subscript 0 stands for the initial state. This becomes
1/2 mv02 + 0 = 0 + mgh
1/2 mv02 + 0 = 0 + mgh
Thus,
h = v02 / 2g = (9.8 m/s)2 / 2(9.8 m/s2) = 4.9 m
The mass of the vaulter cancels out.
