Kinematics - free fall/projectile

  • Thread starter Thread starter rhodium
  • Start date Start date
  • Tags Tags
    Kinematics
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 7K views
rhodium
Messages
9
Reaction score
0

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:
Physics news on Phys.org
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.