Please stop cross-posting to Vortex and other groups.
Prometheus Effect wrote:
>Guys,
>
>How would you calculate the final KE of a vertically
>falling ball assuming you know the mass of the ball, g
>and you could accurately measure the transit time of
>the last say 25mm of the vertical drop?
>
>Then, assuming you knew the total drop distance, would
>not a lower measured KE (than PE theory would predict)
>indicate the amount of effective reduction in the
>acceleration of gravity due to magnetic dragback?
Given
a = acceleration = g
t0 = time at which the first marker is passed
t1 = time at which the second marker is passed
Dt = (t1 - t0)
x0 = height of first marker, x increases DOWNWARD
x1 = height of second marker, x increases DOWNWARD
Dx = (x1 - x0)
v0 = velocity when the ball passes x0
v1 = velocity when the ball passes x1
You can measure t0 and t1, you know x0 and x1, and you want to compute
v1. Right? But v0 will be almost as good (you can compute v1 from it)
and that's what I'll do first.
One more thing; I will define:
I(...)dx == integral(...)dx, 'cause I haven't got an integral sign
here.
Now, start the clock at time t0, so that t0 == 0. Then:
v = v0 + at (for all values of t >= t0)
x1 = x0 + I(v)dt = x0 + I(v0 + at)dt {integral from t0 to t1}
= x0 + v0*Dt + [(a/2) t^2]{from t0 to t1}
= x0 + v0*Dt + (a/2)(t1^2 - t0^2)
Dx = v0*Dt + (a/2)*t1^2
v0 = (Dx - (a/2)*Dt^2)/Dt
v0 = (Dx/Dt) - (a/2)*Dt
Or at any rate that's what I got just now using a bit of scrap paper.
Just measure x0, x1, t0, and t1, plug them in, and you get v0. Then
to find v1,
v1 = v0 + a*Dt
But don't build any bridges based on this unless you double-check it
:-)
