Kinematics - free fall/projectile

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The discussion revolves around solving a kinematics problem involving two objects: one thrown upward with an initial velocity and another falling from a height with zero initial velocity. The equations used for the falling object and the upward-moving object were initially incorrect, leading to confusion about their positions at the time of collision. Participants clarified that the correct approach involves expressing both positions in terms of time and ensuring consistent signs for displacement. The final suggestion was to isolate time in one equation and substitute it into the other to find the collision position. This collaborative effort helped the original poster gain clarity on the problem-solving process.
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Homework Statement


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)


Homework Equations



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:
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rhodium said:
-x= -g/2t^2

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

Hi rhodium! :smile:

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 :wink:)

and the other one is … ? :smile:
 
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:
rhodium said:
ok, so, for equation 2 (object going up), would it be like this:

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

Hi rhodium! :smile:

Yes, if x = h - gt2/2, then y = vt - gt2/2. :smile:
I isolated for x.
x=h-gh^2/(2v^2)

:confused:

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

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