# Height at which objects will collide

1. Sep 22, 2008

### 05holtel

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

A rubber ball is shot straight up from the ground with speed Vo. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.

1) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of Vo, h, and g.

2) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?

3) For what value of h does the collision occur at the instant when the first ball is at its highest point?

2. Relevant equations

Vf^2 = Vi^2 + 2a(h)

3. The attempt at a solution

I am not sure how to make make the symbolic expression

Ball 1:

Vf^2 = Vi^2 + 2a(h)
h= (Vf^2 - Vo^2) / 2a
h= (Vf^2 - Vo^2) / 2(-9.8m/s)
Ball 2:

Vf^2 = Vi^2 + 2a(h)
h= (Vf^2 - Vo^2) / 2a
h= (Vf^2 - 0) / 2(9.8m/s)

Then make H=H

(Vf^2 - Vo^2) / 2(-9.8m/s) = (Vf^2 - 0) / 2(9.8m/s)

2. Sep 22, 2008

### tiny-tim

Hi 05holtel!

This can't possibly work

Hint: you need an equation for h that doesn't involve Vf.

3. Sep 22, 2008

### 05holtel

Hey Tiny Tim,

OK So I tried the formula
sf=si +vi(t2-t1) + 1/2(a)(t2-t1)^2

I get:

Ball 1:
sf = 0 +v0(delta t) + 0.5(-9.8)(delta T)^2
= vo(delta t) - 4.9(delta t)^2

Ball 2
sf = h + (0)(delta t) + 0.5(9.8)(delta t)^2
= h + 4.9(delta t)^2

vo(delta t) - 4.9(delta t)^2 = h + 4.9(delta t)^2

I am not sure what to do because the answer should be in a symbolic expression in terms of Vo, h, and g.

4. Sep 22, 2008

### cepheid

Staff Emeritus
Try this. Let y1 represent the height of the first ball, and y2 the height of the second ball. (We are using the letter y because that is the direction in our coordinate system along which all the movement takes place).

at ANY time t,

$$y_1 = v_{0}t - \frac{1}{2}gt^2$$

because the first ball initially starts off at yi1 = 0 and is thrown up with velocity v0, but begins decelerating immediately due to gravity.

Similarly, at ANY time t, the position of the second ball is given by:

$$y_2 = h - \frac{1}{2}gt^2$$

because the ball starts out at yi2 = h and is just dropped, with NO initial velocity. It immediately begins accelerating downward due to gravity.

At the collision point, clearly the positions of the two balls are the same:

y1 = y2

Solve for the time t at which this occurs, and then plug that time t into either formula to find the height at which it occurs.

5. Sep 23, 2008

### tiny-tim

Hi 05holtel!

(have a delta: ∆ and a squared: ² )

Yes, that's perfect … you got the same equations as cepheid

except that you wrote 9.8 instead of g (which you shouldn't have done, since the question asks you to keep g in the asnwer )

You have: vo(∆t) - g(∆t)² = h + g(∆t)² …

just move it around until you have a quadratic equation in ∆t,

and then solve for ∆t.

("symbolic" just means "no numbers")

(btw, you could have used t instead of ∆t, like cepheid, so long as you specified that the initial time is 0 )