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# Mile High Club

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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Photo attributed to Marquis Lewis.

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### 2 Responses to “Mile High Club”

1. Chris Raabe February 16, 2012 at 2:29 pm #

Can we not assume the 9.8 horizontal velocity becomes a 9.8 vertical velocity, since, as the question states, all kinetic energy is transfered into potential? This would let us use delta v = at, and get a time of one second. In turn, we can apply this one second into the long equation and solve for distance. Delta x = vit + 1/2at^2, giving us 14.7 meters. No?

• MCAT Question of the Day February 18, 2012 at 9:48 pm #

Great analysis and yes this is another way to do the problem. However, what we must remember is direction. If we take 9.8 m/s to be the initial velocity, then we must define gravity as -9.8 m/s^2. So the real equation is:

Delta x = vit – 1/2at^2 = 4.9 m.

Great question!