Calculating the Velocity of a Fireman Sliding Down a Pole

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The discussion focuses on calculating the speed of a fireman sliding down a pole using principles of energy conservation. The fireman starts from rest, and his kinetic energy at the bottom equals the potential energy lost from height h. The equation derived shows that the speed v when the fireman is at height y is v = (2g(h - y))^0.5. A further simplification leads to the expression v = (2g(2h - d))^0.5 when substituting y with d - h. The calculations emphasize the relationship between gravitational potential energy and kinetic energy in this scenario.
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Homework Statement



A fireman of mass (m) slides a distance (d) down a pole. He starts from rest. He moves as fast at the bottom as if he had stepped off a platform a distance h) <(or equal to) (d) above the ground and descended with negligible air resistance.

In terms of (g), (h), and (d), what is the speed of the fireman when he is a distance (y) above the bottom of the pole?

Homework Equations



KE=1/2mv^2
PE=mgh

The Attempt at a Solution



I tried setting KE=PE and making height equal to: h/d-y, but that didn't work. Any ideas? Thanks.
 
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Total Energy remains constant so:

KE_Top + PE_Top = KE_y + PE_y
0 + mgh = 0.5mv^2 + mgy
mg(h-y)=0.5mv^2
2g(h-y)=v^2
v=(2g(h-y)^0.5

y=d-h
Therefore
v=(2g(h-(d-h))^0.5
=(2g(2h-d))^0.5

Is my best bet.
 
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