# Vertical Component of a ball rolling off table

1. Oct 14, 2009

### Chuck Norris

1. The problem statement, all variables and given/known data
A ball rolls off a table with a horizontal velocity of 5 m/s. If it takes 0.6 seconds for it to reach the floor what is the vertical component of the ball's velocity just before it hits the floor? (use g= 10 m/s^2)

The second part of the question is what is the horizontal componenet ofthe ball's velocity just before it hits the floor?

2. Relevant equations

D vert: 1/2at^2

3. The attempt at a solution

I'm not sure if this is correct:

Dvert = 1/2at^2
1/2(10m/s^2)(0.6s)^2
Dvert= 1.8m

For the second part, would I do the exact same thing but plug in different times to plot the different spots the ball would project at on the way down?

2. Oct 14, 2009

### Staff: Mentor

You're supposed to find the velocity, not the distance fallen.

3. Oct 14, 2009

### DaveC426913

2] Seems to me the answers are pretty simple. The vertical component and horizontal component are completely independent of each other. It would seem to me that the horizontal component is trivial (almost a trick question) - it does not even need to be calculated.

4. Oct 14, 2009

### Chuck Norris

So then am I using the wrong formula all together?

5. Oct 14, 2009

### Staff: Mentor

Yep. You're using a formula for distance, but you need the formula for velocity.

6. Oct 14, 2009

### Chuck Norris

V = V0 + at

V= 5 m/s + (10m/s^2)(0.6s)

V= 5.6 m/s?

I'm sure this is the wrong formula also.

7. Oct 14, 2009

### Staff: Mentor

This is the right formula, but you are using the wrong initial velocity. (5 m/s is the horizontal component of the initial velocity, not the vertical.) Treat vertical motion separately from horizontal. (Also: Double check your arithmetic.)

8. Oct 14, 2009

### Chuck Norris

V = Vo + at

V= 0 + (10 m/s^2)(0.6s)

V= 6 m/s

I'm assuming since there is not a initial velocity given in the problem that the initial velocity is assumed to be 0. Sorry for all the dumb questions, physics has just been rough for me so far. Thanks for all your help so far.

9. Oct 14, 2009

### Staff: Mentor

Perfect. The vertical component is 6 m/s downward. (So they might want you to call it - 6 m/s.)

10. Oct 14, 2009

### Chuck Norris

Awesome! Thank you so much. Now for the second part,
what is the horizontal componenet ofthe ball's velocity just before it hits the floor?

Would I use the formula d(horiziontal)=Vot?

Now I'm confused because if I use that formula and plug in the initial velocity of 0 obviously I'm going to get an answer of 0.

11. Oct 14, 2009

### DaveC426913

What is the initial horizontal velocity of the ball? What forces are acting on the ball horizontally (i.e. acceleration)?

12. Oct 14, 2009

### Chuck Norris

The intial horizontal velocity is 5 m/s when it rolls off the table.

So would it be:

d=Vot

D= (5m/s)(0.6s)

d= 3m?

13. Oct 14, 2009

### Chuck Norris

nevermind i see that i'm looking for velocity again yet i went back to answering in distance

14. Oct 14, 2009

### Chuck Norris

Would I just use the same formula as I used with the vertical component but plug in the 5m/s as the initial velocity instead of 0?

15. Oct 14, 2009

### Chuck Norris

Haha wait so is the answer just 5 m/s? Because horizontal velocity doesn't change?

16. Oct 14, 2009

### DaveC426913

You're overthinking it. What is the initial horizontal v? What force(s) are acting to change the horizontal v?

17. Oct 14, 2009

### Chuck Norris

The initial horizontal velocity is 5 m/s correct? I don't believe any forces are acting to change it....

18. Oct 14, 2009

### Chuck Norris

unless the initial horizontal velocity is just zero...since there are no forces changing it

19. Oct 14, 2009

### DaveC426913

Correct and correct.

So...

20. Oct 14, 2009

### Chuck Norris

The answer is just 5 m/s.