# Collision, time

1. Sep 1, 2007

### Niles

1. The problem statement, all variables and given/known data
A ball is thrown straight up from the ground with speed v_o . At the same instant, a second ball is dropped from rest from height H - there's no air resistance.

Find the time when the two balls collide.

3. The attempt at a solution

The path of the ball which is thrown up, B_up, can be described with x_up(t) = ½*(-g)*t^2+v_0*t

The path of the ball which is dropped, B_down, can be descriped with H = ½gt^2

Where do I go from here?

If i assume that the distance the ball which is thrown up travels is H, then H = ½*(-g)*t^2+v_0*t, and inserted in H = ½gt^2, I get that t = v_0/g. Does this look correct?

2. Sep 1, 2007

### Staff: Mentor

Good. Note that the starting point is the ground, which is x = 0.

Careful. Use the same kind of formula as above, the only difference is the initial position and speed. Make sure that both formulae measure postion relative to the same place. Then you can solve for the time when both are at the same position.

3. Sep 1, 2007

### Niles

H = ½*(-g)*t^2+v_0*t - H = ½*(-g)*t^2 - H ?

4. Sep 1, 2007

### Staff: Mentor

Not sure what you are doing here. Write an expression for the position of the dropped ball (x_down) as a function of time. Its initial position is x_down = H.

5. Sep 1, 2007

### Niles

so x_down(t) = ½*g*t^2+v_0*t + H ?

x_down(t) = H, right? So from here it's x_down = x_up, and find t?

6. Sep 1, 2007

### Staff: Mentor

Almost: Careful with the sign of the acceleration.
x_down(t=0) = H.
Yep!

7. Sep 1, 2007

### Niles

If the acceleration for x_up is -g, then for x_down it must be g?

8. Sep 1, 2007

### Staff: Mentor

Oh really? According to Newton, force and acceleration must act in the same direction. Which way does the force of gravity act? Does it depend on whether an object is rising or falling?

(To measure position consistently, use a uniform sign convention; for example: let up = +, down = -.)

9. Sep 1, 2007

### Niles

If that is so, then why is x_up(t) = ½*(-g)*t^2+v_0*t and x_down(t) = ½*(-g)*t^2+v_0*t + H?

For x_up I say it's -g, and for x_down I say g, but that was wrong?

10. Sep 1, 2007

### Staff: Mentor

That's almost right: Is v_0 the same for both?

Yes, that was wrong. The acceleration due to gravity is always downward, which is -g using our standard sign convention. (The letter "g" always stands for the magnitude of the acceleration due to gravity; g = 9.8 m/s^2.)

11. Sep 1, 2007

### Niles

x_up(t) = ½*(-g)*t^2+v_0*t and x_down(t) = ½*(-g)*t^2+v_0*t + H.

For x_down v_0 is zero - that is what I assume. But none the less, they are not the same.

So for x_up and x_down, it's -g?

12. Sep 1, 2007

### Staff: Mentor

Right: They are not the same. So you'd better change one of your expressions!

Yep.

13. Sep 1, 2007

### Niles

Great. I get t = H/v_initial, up

14. Sep 1, 2007

Looks good.