# Kinematics - free fall/projectile

1. Sep 22, 2008

### rhodium

1. The problem statement, all variables and given/known data
The problem is that basically, if an object is thrown up with a certain initial velocity (v), and an object at height h directly above the first object falls (intial velocity=0), write an equation for the position they collide. (position denoted as x)

2. Relevant equations

3. The attempt at a solution

ok I tried this question so many times (actually about 3 times). This is the equations I used

object 1 (falling down)
-x= -g/2t^2

object 2
h-x= vt - g/2t^2

I isolated them for x, but the answer is wrong. Can anyone tell me what is wrong with my 2 equations?

edited to add: this is what the 2 equations solved to:
x=gh^2/(2v^2)

Last edited: Sep 22, 2008
2. Sep 23, 2008

### tiny-tim

Hi rhodium!

If v is measured up, then s must be measured up also,

so the falling one is x= -gt^2/2 (or x = h - gt^2/2 , depending where you're starting from )

and the other one is … ?

3. Sep 23, 2008

### rhodium

hi,

i am still confused. i put - on the s because the displacement is negative. but sicne the initial position is h, then it would make sense to write it that way. thanks!

ok, so, for equation 2 (object going up), would it be like this:

y=vt-g/2t^2 where x=y?

i hope i got it right.

edited to add:

I isolated for x.
x=h-gh^2/(2v^2)

Last edited: Sep 23, 2008
4. Sep 24, 2008

### tiny-tim

Hi rhodium!

Yes, if x = h - gt2/2, then y = vt - gt2/2.

No … put x = y, and solve for t … then use that t to get x.

5. Sep 24, 2008

### rhodium

all right,

thank you for teaching me. i was really stuck before.

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