A man at the top of a building 20m high releases a stone from rest; 0.60 seconds later he throws a marble vertically downwards with an initial velocity of 8.0 ms-1. How long after the stone was dropped does the marble pass the stone?
I know that after 0.6 seconds the stone is travelling at 6 ms-1, and because the acceleration is constant, this means that the marble is always travelling 2 ms-1 faster than the stone. I'm thinking that I need to make equations for the distance of each object, equate them and solve for t, but I can't seem to get started. If you could point me in the right direction, that would be great.
Just google Newton's equations of motion.
Incidentally I notice an error in the first equation
for the first equation they are the same thing since t0=0 but I should have been more careful. The same is true for the v1(0) there. It should be v1(t0).
I'm thinking that I need to make equations for the distance of each object, equate them and solve for t, but I can't seem to get started. If you could point me in the right direction, that would be great.
All of my textbooks, assignments and exams for the course recognise the acceleration due to gravity as 10m/s2 (which obviously isn't accurate), and as such my physics teacher asks us to use that value in our calculations.
As for the question, assuming gravity to be 10m/s2, I got:
20 - 5t = 20 - 5(t - 0.6) - 8(t - 0.6)
8t = 7.8
t = 0.975 seconds
EDIT: Just noticed that romsek used "a(t - t0)2", whereas HallsofIvy used "(1/2)a(t - t0)2".
Let t=0 when marble is dropped.
At t=0 the stone is at s0s with velocity v0s:
s0s = (1/2)g(.6)2
v0s = g(.6)
Stone: ss = s0s + v0s(t) + (1/2)gt2
Marble: sm = v0m(t) + (1/2)gt2
ss=sm -> t = s0s/(v0m-v0s)
Physically, this says, since marble has same acceleration as stone, marble will never catch up to stone unless its initial velocity is greater than the velocity of the stone at .6 secs.