What is the speed with which she lands?

[kokenwa]

A pole-vaulter of mass 60kg vaults to a height of 6.0m before dropping to thick padding placed below to cushion her fall. What is the speed with which she lands?


-I know it has something to do with momentum, but I don't even know where to begin on this one

 

[rock.freak667]

Consider neglecting air resistance and using the law of conservation of energy

 

[ogb p]

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The speed with which the pole-vaulter lands can be calculated using the principle of conservation of energy. When the pole-vaulter is at the height of 6.0m, she has potential energy due to her position above the ground. As she falls, this potential energy is converted into kinetic energy, which is the energy of motion. At the moment of impact with the padding, all of her potential energy will be converted into kinetic energy.

To calculate her speed, we can use the equation for kinetic energy: KE = 1/2 * m * v^2, where m is the mass of the pole-vaulter and v is her velocity. We can also use the equation for potential energy: PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.

Setting these two equations equal to each other and solving for v, we get:

1/2 * m * v^2 = mgh

v^2 = 2gh

v = √(2gh)

Plugging in the values given in the problem, we get:

v = √(2 * 9.8 m/s^2 * 6.0m)

v = √(117.6 m^2/s^2)

v = 10.84 m/s

Therefore, the speed with which the pole-vaulter lands is approximately 10.84 m/s. It is important to note that this calculation assumes no air resistance or friction, so the actual landing speed may be slightly lower.