# Shot put physics problem

1. Sep 21, 2004

### Blassemer

A shot put is released with a velocity of 12 m/s and stays in the air for 2.0s.

At what angle with the horizontal was it released?

What horizontal distance did it travel?

2. Sep 21, 2004

### Pyrrhus

Some hints:
Consider when y =0 and when y is at max height.

3. Sep 21, 2004

### Blassemer

ok im not good at this stuff at all, lol

can someone plz give me a step by step of it or give me a lil more info on how to get the answer

4. Sep 21, 2004

### HallsofIvy

Staff Emeritus
But the shot put is not at y= 0 when it is thrown. I don't see how you can do this without knowing the initial height of the shot.

(By the way- a "shot put" is NOT released. "Shot put" is the name of the event. It is the "shot" that is thrown (or "put").

5. Sep 21, 2004

### Sirus

I think you need the angle at which it was released. Blassemer, are you sure that's all the information given by the question?

PS: HallsofIvy breaks 3000! :tongue2:

Last edited: Sep 21, 2004
6. Sep 21, 2004

### Pyrrhus

Well, assuming $$Y_{o} = 0$$. I agreed with Halls, but maybe he left it out.

Using this equation

$$Y - Y_{o} = V_{yo}t + \frac{1}{2}at^2$$

$$0 = V_{yo}t_{f} + \frac{1}{2}gt^2_{f}$$

$$t_{f}$$ It's flight time and is equal to 2 seconds.

We can get Initial Vy component that way.

$$V_{o}^2 = V_{xo}^2 + V_{yo}^2$$

$$V_{o} = 12 m/s$$

Last edited: Sep 21, 2004
7. Sep 21, 2004

### Sirus

I don't think that works, because you use the formula $d=v_{i}t+\frac{1}{2}at^2$ for the up and down movement, but that formula can only be used for uniformily accelerated motion. In this case, the vector g changes from positive to negative, so you cannot use that equation, unless I am misunderstanding something.

Last edited: Sep 21, 2004
8. Sep 21, 2004

### Pyrrhus

It looks to me like simple projectile motion, so you can use the uniform acceleration equations. Of course, it's rather ambigous, so i worked the problem on that premise.

Last edited: Sep 21, 2004
9. Sep 21, 2004

### Sirus

Actually, besides my earlier concern, I think your equation work assumes Y, the upward vertical displacement, is 0, which is incorrect. And why did you switch to a minus sign on the second line of you latex? Sorry if I'm not understanding something.

10. Sep 21, 2004

### Pyrrhus

Well if it is projectile motion and you shoot a bullet, and it says it stay in the air for 2 seconds, then that must be the whole flight time. Y=0 because the bullet will had hit ground. My sign convention is down and left negative, and up and right positive. Oh i like to put the sign to the variables, i could have left it as positive and then plug in -9.80 instead of 9.80. I will change it for sake of avoiding confusion.

Last edited: Sep 21, 2004
11. Sep 22, 2004

### HallsofIvy

Staff Emeritus
That's an awfully short shot-putter!

Yeah, I need to get a life!

12. Sep 22, 2004

### Sirus

Ok. But I still don't think you can use that formula for motion that changes direction, because such motion is not uniformily accelerated. Same applies to a bullet: you can only use kinematics formulas for the up or down motion separately, not for the total displacement (which would be zero). I don't think this problem can be done without more information.

13. Sep 22, 2004

### Sirus

Actually, now that I think of it, the acceleration is constant (gravity). But I'm still not sure if that works. I think we've succeeded in seriously confusing Blassemer. Problem can't be done, in my opinion.

14. Sep 22, 2004

### Pyrrhus

I'm spanish, i've no idea what a shot putter is :rofl: . I thought it was like a bullet....

Ok, we all agree we need more information to do the problem, until that is provided, there's no accepted solution, just assumptions. Please Blassemer comply with the petition, so we can help you out.