Distance fallen with velocity-squared drag from v0 to v1

[qqchico]

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

 

[James R]

Are you supposed to use differential equations to solve this?