a depth of mass a00 kg is released from a warship into the sea and allowed to sink. While gravity is pulling it down, a buoyancy force of 1/40 times its weight (=mg) is pulling it up. therefore, water resistance also exerts a force on the weapon that is proportional to its velocity, with a constant of proportionality of 10 kg/s. How long will it take for the weight to reach the velocity of 70 m/s? i've tried and my de equation is : f = ma mg-1/40(mg)-10v = m(dv/dt) dv/dt = g - )1/40)g - (10/m) v dv/dt = (39/40)g - (10/m) v v(t) = (39/40)gv - 5(v^2)/m + c applying the initial condition (x=0, t=o, v=0) => c=0 thus v(t) = (39/40)gv - 5(v^2)/m is my de is right??? how i get t expression from this equation???