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.
Hello,
Welcome to these forum sites, whatever you have asked, I don’t have
right answer to this, but if you want to come India, you come
to the,
*************
India Tour
Thanks,
You can really find this derivation in a thousand places on the web worked out with pictures and animations etc. that I certainly can't show you here.
Just google Newton's equations of motion.
Incidentally I notice an error in the first equation
should read
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).
Are you rounding to one decimal place? Why would you do that? The downward acceleration, due to gravity, is 9.2 m/s^{2} so that after 0.6 seconds, the stone will be travelling at (9.2)(0.6)= 5.52 m/s downward. So the difference in speeds will be 8.00- 5.52= 2.48 m/s. In any case, the height of the stone, t seconds after it was thrown will be -4.6t^2+ 20 while the height of the marble will be -4.6(t- .6)^2- 8(t- .6)+ 20. The marble will "pass" the stone where their heights are equal: -4.6t^2+ 20= -4.6(t- .6)^2- 8(t- .6)+ 20. The t^2 terms will cancel out leaving a linear equation to solve for t.
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/s^{2} (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/s^{2}, 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)gt^{2}
Marble: sm = v0m(t) + (1/2)gt^{2}
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.