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

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To determine the necessary height for a shot tower where ball bearings solidify in 4 seconds, the acceleration due to gravity (9.8 m/s²) is used in kinematic equations. The height can be calculated using the formula y = 1/2 * a * t², resulting in a height of approximately 78.4 meters. The impact velocity of the bearings can be calculated as well, which is essential for understanding the process. The discussion highlights the importance of correctly applying physics principles and equations to solve the problem. Ultimately, the correct calculations lead to the necessary height and impact velocity for the shot tower.
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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??
 
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Lida said:
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??
Welcome to PF Lida,

What is the acceleration of the shot?
 
Hootenanny said:
Welcome to PF Lida,

What is the acceleration of the shot?


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


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

And thanks for the welcome. :)
No problem, I see from your initial post that this is your first physics class, do you have a class text? Have you come across kinematic (SUVAT) equations before?
 
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!
 
Lida said:
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?
Correct. 9.8 m/s2 is the acceleration due to gravity or the acceleration of an object in freefall.
Lida said:
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?
Not quite, perhaps if you detailed you calculations we could point out where you're going wrong.
 
Lida said:
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!

No, that's incorrect. Use this equation, and solve for y (height). Make sure that you plug in your known variables: a (acceleration), t (time).

y = \frac{1}{2}at^2
 
chislam said:
No, that's incorrect. Use this equation, and solve for y (height). Make sure that you plug in your known variables: a (acceleration), t (time).

y = \frac{1}{2}at^2

With this formula I got 192.08m for the tower and 48.02m/s for my velocity. Closer?
 
Hootenanny said:
Not quite, perhaps if you detailed you calculations we could point out where you're going wrong.

I multiplied 9.8 by 4 in an attempt to reverse the equation for acceleration.
 
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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 x-x0= 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?
 
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  • #12
I think the answer is 78.4 now...hopefully.
 
  • #13
Lida said:
I think the answer is 78.4 now...hopefully.
Correct :approve:
 
  • #14
Finally!

Thanks! :biggrin:
 

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