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Two objects in free fall

  1. Mar 30, 2016 #1
    1. The problem statement, all variables and given/known data

    An inquisitive physics student and mountain climber climbs a 50.0m high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously? (c) What is the speed of each stone at the instant the two stones hit the water?

    2. Relevant equations
    X=Xo-0.5gt^2

    3. The attempt at a solution
    a) X=-2t-4.9t^2
    50=-2t-4.9t^2
    t=2.99s

    (b) X=V(t-1)-4.9(t-1)^2...

    The thing that I dont understand is that the question states that both stones hit the ground at the same time (splash together), but in solving this problem we use "t-1" which would mean that the second stone hits the ground one second before the first stone. Am I interpreting the question wrong. Could you please explain fully
     
  2. jcsd
  3. Mar 30, 2016 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    The 2nd stone is thrown 1 second after the first, but with a higher initial vertical velocity. That is how it can catch up to the first stone just as they reach the water. Does that help?
     
  4. Mar 30, 2016 #3
    I understand what you are pointing out. But im confused if both stones do touch the ground at the same time because the second stone take 1.99s to touch ground while the first takes 2.99s
     
  5. Mar 30, 2016 #4

    berkeman

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    Staff: Mentor

    That matches them being thrown 1 second apart, no? :smile:
     
  6. Mar 30, 2016 #5
    So the time at which they reach the ground is not equal?
     
  7. Mar 30, 2016 #6

    berkeman

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    Staff: Mentor

    They reach the surface of the water at the same time Tfinal. They are thrown at two different starting times. Their flight distances are equal. That's how you solve for the actual flight time durations...
     
  8. Mar 30, 2016 #7
    Ok thanks alot I think I understand a little bit. Really appreciate your help. God bless you
     
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