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Free fall question

  1. Jun 12, 2007 #1
    1. The problem statement, all variables and given/known data

    In part (a) of this question suppose a stone is thrown verically upward with a speed of 12.9 m/s from the edge of a cliff and that h = 82 m. At what speed (in m/s) does the stone hit the ground?

    2. Relevant equations

    T = v/g
    Free fall equation
    V = (V^2 - 2gvt . sin(feta) + (g^2) . (t^2))^0.5

    3. The attempt at a solution

    By using the first equation, I have worked out how long it takes to reach max. Then by using the answer for that, and using the known values, I have used the free fall equation to find out how long it takes to reach the bottom of the cliff. Then by using the last equation I ended up with 52.956 m/s as the answer.

    I was told this was not the answer and was wondering what steps I am missing or completely doing wrong.

    The question may sound silly but I guess what its trying to ask is 'what is the velocity of the stone right before it hits the ground?'.

    Any help would be most appreciated.
  2. jcsd
  3. Jun 12, 2007 #2
    wow! You do not need to use such a complicated equation!

    The rock is thrown up. It would soon change direction to start coming down. Calculate this height by using v^2=u^2-2as (by solving for s, also v=0 obviously).

    Then add this calculated s to 85. Then use the same equation to figure out v. If you need further help ask away...:)
  4. Jun 12, 2007 #3
    Sorry, but where are you getting 85 from?
  5. Jun 12, 2007 #4
    Okay, there is a misunderstanding, may you please rephrase this question? I thought, h was the height of the cliff
  6. Jun 12, 2007 #5
    Ah, height is part of the question, but I guess you meant 82, you wrote down 85. And what do you mean by add s (seconds I presume) to 82 (height)?
  7. Jun 12, 2007 #6
    Oh sorry, Anyway, can you see how it works?
  8. Jun 12, 2007 #7
    Not really. For starters, u = height?, and could you please refer to my previous post for the second bit I don't understand.
  9. Jun 12, 2007 #8
    Okay, one of the 4 basic equations of motion is:
    v=final velocity
    u=initial velocity

    Now we want distance.
    so we solve for s.
    therefore: s=(v^2-u^2)/(2g)

    Okay. When the ball reaches its maximal height, its velocity is 0 (Once you can accept that any projectile question becomes a poece of cake). So that is why we set:
    g=-9.8 (the object is DECELERATING due to the effect of gravity...)

    Okay do you get it so far? So work out s and find the distance from the ground to the maximal height of the rock (not from the cliff).
  10. Jun 12, 2007 #9
    I see that you have used time. Try my method using distance.
  11. Jun 12, 2007 #10
    Ok, so using that equation, I have gotten distance = 8.49m

    Is that the maximum height the stone reaches? If so, where do I go from here to obtain the speed of the stone just before it hits the ground?
  12. Jun 12, 2007 #11
    Good. So the distance from the ground (not cliff) is 8.49+h = 90.49

    Okay. Now use v^2=u^2+2as
    Here v is what we want to find.
    Now is there an initial velocity? No. It is 0.
    a=+9.8 (the stone/rock is ACCELERATING)
    well I think the answer can be found now.

    So remember acceleration whether it is a minus or plus. And try to find the easier equation to use in the context.
  13. Jun 12, 2007 #12
    So therefore,

    v = (2x9.8x90.49)^0.5

    = 42.11 m/s

    I'm fairly certain that is the answer. Thankyou very much for being so patient with me and helping me out, its my first time doing physics.

    On a side note, there is another I need a tiny bit of help on. Got the first part of this question, but the second part I just can't grasp.

    Part A

    A cheetah, the fastest of all land animals over a short distance, can accelerate from zero to 18.0 m/s in three strides and to a full speed of 31.3 m/s in seconds. Assuming the first three strides are each 4.6 m long and that acceleration is constant until the cheetah reaches full speed, what is the cheetah's acceleration (in m/s2)?

    The answer is 11.88 m/s2.

    Part B

    If in part (a) of this question the cheetah's first three strides were each 3.7m long, how many seconds would it take to reach its full speed?

    I keep saying to myself the equation I should be using is S = D/t
    S = velocity/speed
    D = distance/displacement
    t = time

    When it says the strides are now 3.7m long, I'm guessing a new acceleration is required, so I've come up with 14.77 m/s2. The answer is 2.14 seconds, but I just can't work out the solution. (Answer can be within .1).
  14. Jun 12, 2007 #13
    That is wierd. Show how you got the answer to part a
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