A pole vaulter vaults over the bar and falls onto thick padding. He lands with a

Question

A pole vaulter vaults over the bar and falls onto thick padding. He lands with a speed of 9.8m/s; the padding then brings him to a stop in a time of 0.40s. What is the average force on his body due to the padding during that time interval? Express your answer as a fraction or multiple of his weight. [Hint: The force due to the padding is not the only force acting on the vaulter during the 0.40s interval.] (Answer in the textbook came up with: 3.5 times his weight) Trouble figuring out how they came up with that.

Explanation / Answer

The padding must decellerate an object moving 9.8 m/s + (0.4)(9.8)m/s in 0.4 sec. That is, he has reached a velocity of 9.8 m/s at the instant of contact and is still accelerating under the influence of gravity at 9.8 m/sec^2 for an additional 0.4 sec until he is stopped. So, the vaulter reaches an "effective" velocity of 9.8 m/s + 3.9 m/s = 13.7 m/s while being decellerated at a rate of (13.7m/s)/(0.4s) = 34.3 m/s^2. The decelleration rate in "g's" (or multiples of his weight) is therefore: 34.3 m/sec^2 / 9.8 m/sec^2 = 3.5 g

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