Asky diver steps from a high-flying helicopter. if there were not air resistance, how fast would she be falling at the end of a 12 second jump?

If there is no air resistance and the body, in this case the sky diver simply falls from the helicopter then, the motion can be described as a free-falling body. The equation that would allow us to determine the speed of the body at any time, t is,

                                    v(t)  = vo + gt

where v(t) is the unknown velocity, vo is the initial velocity which is equal to zero and g is the acceleration due to gravity which is equal to 9.8 m/s2.

Substituting the known values,

                                    v(t) = 0 + (9.8 m/s2)(12s)

                                                v(t) = 117.6 m/s

Hence, the speed of the sky diver at the end of 12 second is approximately 117.6 m/s.  

your answer is, orbit around the nucleus