# Height at which objects will collide

## Homework Statement

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?

## Homework Equations

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

## 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)

tiny-tim
Homework Helper
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)

Hi 05holtel! This can't possibly work Hint: you need an equation for h that doesn't involve Vf. 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.

cepheid
Staff Emeritus
Gold Member
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.

tiny-tim
Homework Helper
I get:

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.

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 )