Thread: initial conditions the hammer and the feather

A 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.....

d2s/dt2=-5.2ft/sec^2

initial conditions: ds/dt=0 and s=4 when t=0.

So, I find ds/dt, is -5.2T+c, and v(0)=4.

So v(t)=-5.2t+c=0

now s(t)=(-5.2/2)*t^2+cx+x

I want to find t when S(t)=0

4=(-5.2/2)0^2+kt+c
c=4

I want t when s(t)=0.

0=(-5.2/2)t^2+kt+4

what now?

Originally Posted by StudentMCCS

A 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.....

d2s/dt2=-5.2ft/sec^2

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

corrections made