corrections madeA hammer and a feather are dropped at the same time from 4 feet, in a under a vacuum by an astronaut under the moons gravity.....
initial conditions: ds/dt=0 and s=4 when t=0.
So, I find ds/dt, is -5.2T+c, and v(0)=4. no, v(0) = ds/dt when t = 0
So v(t)=-5.2t+c=0 v(t) = -5.2t
now s(t)=(-5.2/2)*t^2+cx+x s(t) = -2.6t^2 + C
s(0) = 4
4 = -2.6(0)^2 + C
s(t) = -2.6t^2 + 4