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How do I find the necessary height for a shot tower in this problem? 
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#1
Sep408, 07:18 PM

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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? I've never taken a physics course before. I have no idea how to figure out the height of the shot tower, I think I'm supposed to know from the 4 seconds that the tower is a certain height, but I'm totally clueless. If someone could even just tell me how to figure that out I'd really appreciate it. I saw an old post with a near identical question from 2 years ago, but looking at the answer given there didn't help me figure out what to do here. Help?? 


#2
Sep408, 07:29 PM

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What is the acceleration of the shot? 


#3
Sep408, 08:03 PM

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I dooooon't knooooow! :( Should it be obvious from the problem? I'm so confused by this. And thanks for the welcome. :) 


#4
Sep408, 08:06 PM

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How do I find the necessary height for a shot tower in this problem?



#5
Sep408, 08:37 PM

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Ok. I have a book called "How to Solve Physics Problems" and it uses 9.8 a lot for free falling objects, so I'm guessing that might be the acceleration of a falling object?
If that's the case, then the shot tower is 39.2 m right? And I got that the velocity was 9.8 m/s, but I'm not so sure about that. Is it right? Thanks! 


#6
Sep408, 08:40 PM

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#7
Sep408, 08:42 PM

P: 79

[tex]y = \frac{1}{2}at^2[/tex] 


#8
Sep408, 08:50 PM

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#9
Sep408, 08:52 PM

P: 10




#10
Sep408, 08:56 PM

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#11
Sep408, 09:19 PM

P: 10

Now I figured 39.2 as my velocity and got 348.88 for the tower. This doesn't seem right, but I did that using the xx0= v0t + 1/2at^2 equation. ( I'm using 192.08 for 1/2at^2)
Using 19.6 instead of 192.08, because I think that might be where I'm going wrong, I end up with 176.4. Are either of my answers right? How do I figure out the difference between x and x0? 


#12
Sep408, 09:42 PM

P: 10

I think the answer is 78.4 now...hopefully.



#14
Sep508, 06:46 AM

P: 10

Finally!
Thanks! 


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