Recent content by basketball5rya

Alright I went ahead and tried to differentiate equation. m*dv/dt=mg-kv dv/dt=g-(kv/m) dv=(g-(kv/m))*dt dt=(1/(g-(kv/m))*dv Let: u=g-(kv/m) du=(-k/m)dv dv=(-m/k)du (-m/k)du/u=dt (1/u)du=dt(-k/m) ln(u)=(-k/m)t+C e^ln(u)=e^[(-kt/m)(kC/m)] g-(kv/m)=e^[(-kt/m)(kC/m)] v=(m*e^[(-kt/m)(kC/m)]-g)/-k...

I have in calculus but not in physics before. Was your first reply correct? The F=mg-FR=mg-kv? If so, I got v=1/2 m/s. F=mg-kv ma=mg-kv 3(9)=3(10)-6v -3=-6v 1/2=v I guess I need a full on explanation on what's going on in this problem :/ sorry

Resistance -- fluid resistance to ball motion A ball of mass m = 3 kg is falling from an initial height y = 3 m in an area where the acceleration due to gravity can be approximated as g = 10 m/s2. As it falls the ball is subjected to a fluid resistance of magnitude FR = kv where k = 6 kg/s. How...