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 v
not to v
1 is given by
s(v
not-> v
1)= 1/2 [(g-kv
not^2)/(g-kv12}]
I hate proofs
