If you dropped two identical balls from rest, one at t=0 and the other two seconds later?

Question

Both of them will eventually reach same terminal velocity v = 75 m/s.

What will be eventual vertiacal separation of the balls?

Answer 1

Both balls will reach V = 75 m/sec at the same place in their respective falls...call it D. This results because D = f(a(t),v(t)), which is a function (f) of a(t) and v(t); where a(t) is the time since drop dependent acceleration of each ball as it accelerates to V and v(t) --> V, also time dependent, as each one accelerates. Thus a and v are the same at the same time since drop (t) for both balls.

But the second ball will reach the same D two seconds later than the first ball because it remains two seconds behind (delt = 2 seconds) throughout the fall. Meanwhile, when ball two reaches D, ball one will have traveled S = Vdelt = 75*2 = 150 meters at a constant terminal velocity.

Once ball two reaches D, both balls are at the same V = 75 m/sec and remained fixed at that speed. So their separation remains at 150 meters until ball one crashes and comes to an abrupt halt.

For this problem, D is the zero point and we really don't care what the absolute value of D is. So this is like a foot race, but ball one starts two seconds earlier than ball 2 from the starting block at D.

Answer 2

it will be two seconds behind still .. and how far is that is they are both travelling at 75m/s?

:)