# Someone give me a hint been at it for hrs

1. Sep 26, 2009

### sjcc

1. The problem statement, all variables and given/known data

A ball is kicked from a location < 7, 0, -6 > (on the ground) with initial velocity < -9, 15, -3 > m/s. The ball's speed is low enough that air resistance is negligible.
What is the velocity of the ball 0.4 seconds after being kicked? (Use the Momentum Principle!)

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 26, 2009

### Pengwuino

What are your thoughts on the problem? :)

3. Sep 26, 2009

### sjcc

i've tried dividing the location of the ball by the 0.4 seconds. it was wrong. I thought of next to multiply the location by 0.4. but Im not sure if that is correct. i think there is something that needs to be done about "The ball's speed is low enough that air resistance is negligible".

4. Sep 26, 2009

### Pengwuino

The fact that air resistance is negligible simply states that there is no friction. Dividing a location by a time is meaningless, as is multiplying a location by a time. Your velocity is of the form $$(v_x,v_y,v_z)$$ where the vertical direction is y. If your only under the influence of gravity, how does $$v_y$$ change over time?

5. Sep 26, 2009

### sjcc

LaTeX Code: v_y should decrease over time.

6. Sep 26, 2009

### Pengwuino

Yes, and in a gravitational field, you know it decreases by 9.8m/s^2. Thus, you can determine what the $$v_y$$ velocity is after the time given.

The question might want a different method though, what does the question mean by momentum principle?

7. Sep 26, 2009

### sjcc

pfinal = pintial + the net force * the change in time.

8. Sep 26, 2009

### Pengwuino

Well, that seems a little silly to use that method but ok. Remember, $$p = mv$$ so $$p_f = mv_i + F_{net} * \Delta t$$. You know the force of gravity on an object is $$F = mg$$ where g = -9.8m/s^2. Thus, you can easily determine the final momentum and thus, final velocity for the y-direction. Of course, the x and z velocities do not change as they are not affected by gravity.

9. Sep 26, 2009

### sjcc

in the problem, there is no M stated.

10. Sep 26, 2009

### Pengwuino

Just say m = M, it will cancel out since the mass does not change

11. Sep 26, 2009

### sjcc

thanks. i was able to figure out the prob. j/w can you clear me up on graphs of motion? i am a bit confused on how to read them.

12. Sep 26, 2009

### Pengwuino

What are you having problems with inregards to them?