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

1. Sep 4, 2008

### Lida

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. Sep 4, 2008

### Hootenanny

Staff Emeritus
Welcome to PF Lida,

What is the acceleration of the shot?

3. Sep 4, 2008

### Lida

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

And thanks for the welcome. :)

4. Sep 4, 2008

### Hootenanny

Staff Emeritus
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?

5. Sep 4, 2008

### Lida

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. Sep 4, 2008

### Hootenanny

Staff Emeritus
Correct. 9.8 m/s2 is the acceleration due to gravity or the acceleration of an object in freefall.
Not quite, perhaps if you detailed you calculations we could point out where you're going wrong.

7. Sep 4, 2008

### chislam

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).

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

8. Sep 4, 2008

### Lida

With this formula I got 192.08m for the tower and 48.02m/s for my velocity. Closer?

9. Sep 4, 2008

### Lida

I multiplied 9.8 by 4 in an attempt to reverse the equation for acceleration.

10. Sep 4, 2008

### Hootenanny

Staff Emeritus
Last edited by a moderator: Apr 23, 2017
11. Sep 4, 2008

### Lida

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?

Last edited: Sep 4, 2008
12. Sep 4, 2008

### Lida

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

13. Sep 5, 2008

### Hootenanny

Staff Emeritus
Correct

14. Sep 5, 2008

Finally!

Thanks!