Help with Proof: Particle Falling Distance from vnot to v1

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The discussion centers on deriving the distance a particle falls under the influence of gravity and a resisting force proportional to the square of its velocity. The formula established is s(vnot -> v1) = 1/2 [(g - kvnot^2) / (g - kv1^2)], where g represents gravitational acceleration and k is the proportionality constant for the resisting force. The participants confirm that differential equations are necessary to solve this problem, as they provide the framework for modeling the motion under the given forces.

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qqchico
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the question asks consider a particle of mass m whose motion starts from rest in a constant gravitational field. if a resting force proportional to the square of the velocity (i.e, kmv^2) is encountered, show that the distance s the particle falls from vnot to v1 is given by

s(vnot-> v1)= 1/2 [(g-kvnot^2)/(g-kv12}]

I hate proofs :eek:
 
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Are you supposed to use differential equations to solve this?
 

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