Solving Ball Bearing Problem: Tower Height & Impact Velocity

In summary, the conversation discusses the process of making ball bearings using a shot tower and the equation used to determine the height of the tower and the impact velocity of the bearing. The equation used is Δx = .5(g)(t)2, which can be derived from the general equation for accelerated motion. The conversation also mentions the difficulty of typing equations and finding the correct values for s0 and v0 in this scenario.
  • #1
Hi, I was Googling about the following question :

"Ball bearings can be made by leting shperical drops of molten metal fall inside a tall tower - called a shot tower- and solidify as they fall.

If a bearing needs 4.0s to solidify enough for impact, how high must the tower be?
What is the bearing's impact velocity?"

In order to find some information on how to do it, and came across this forum and found another post about it that has the equation:
Δx = .5(g)(t)2

My problem is that I've gone through my book making a formula sheet of all the formulas from the chapters we've covered and I'm having a problem deriving this formula from other formulas and can't seem to come up with a solution in how to come to this equation. Can someone possibly help me with this?

I did get the answer (from yahoo using the same formula). However, later I did some maneuvering of my own using the definition of acceleration as the rate of change of velocity (ay=(ΔVy/Δt)) to find ΔVx, and then plugged ΔVy into the equation for relating velocity and displacement for constant-acceleration motion ((Vy)f2 = (Vy)i2 +2ayΔy), and with some algebraic maneuvering found the same answer (78.4m). The previous equation seems far more simple though.

Hopefully my equations came out correct.
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  • #2
Doing all those super and sub-scripts are difficult.
  • #3
v = a*t

Integrate once.
  • #4
The general equation for accelerated motion is:

s(t) = s0 + v0*t + (1/2)*a*t^2

For your shot tower, what are s0 and v0?
  • #5
They would be zero. I see how the equation is found now.

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