1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Constant acceleration problem

  1. Nov 5, 2007 #1
    1. The problem statement, all variables and given/known data

    I have 2 rocks. One is thrown straight up from a cliff's edge. A certain amount of time later another rock is dropped from the cliffs edge.

    I need to calculate the initial velocity of the rock 1 so that they hit the ground simultaneously. I have the cliff hight, the time after rock 1 when rock 2 was thrown, and acceleration of course.


    I just need some pointers as I can't seem to get my thinking behind it.

    Cheers!!!!
     
  2. jcsd
  3. Nov 5, 2007 #2
    The only thing that is going to change is the initial velocity of the thrown rock. Which is what we're looking for. Would you agree that the dropped rock with always hit the ground in the same time interval since the height initial velocity and acceleration of gravity wont change?

    Does this help at all?
     
  4. Nov 5, 2007 #3
    Yes I agree with that. I have the time it takes to hit the ground from s=ut+(1/2)at^2.

    The first rock is in the air for this time plus the time interval when the 2nd rock started. Correct? Is this needed?
     
  5. Nov 5, 2007 #4
    Exactly correct! That's how much time the thrown rock will be in the air total. Do know how to go about solving from here?
     
  6. Nov 5, 2007 #5
    Mmm....I'm not exactly sure yet. What confuses me is that it is going up and decelerating. Then it accelerates down again a distance the same as rock 2 plus and unknown distance.
     
  7. Nov 5, 2007 #6
    Ah, do you know the equations of motion for constant acceleration?
     
  8. Nov 5, 2007 #7
    v=u + at

    s=ut + (1/2)at^2

    v^2=u^2+2as


    Are they relevant here? Maybe this is all really easy, it's probably so easy but my brain is tired!!
     
  9. Nov 5, 2007 #8
    Sure we don't the height in flight but s=ut + (1/2)at^2 doesn't require it does it? You know the initial position the rock was thrown and you know where it lands correct?

    "s" is the final position of the rock, if we call the cliff 0 then the final position, the ground, is a negative amount of units below the cliff right?
     
  10. Nov 5, 2007 #9
    Right....?
     
  11. Nov 5, 2007 #10
    Ok, I got it!!


    I realised that distance above the cliff is negligible because it's displacement is 0 back at the cliff when it comes down.....
     
  12. Nov 5, 2007 #11
    Yup! Nice
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Constant acceleration problem
Loading...