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Another aggravating kinematics problem - regarding stones

  1. Sep 21, 2006 #1

    OVB

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    A stone is DROPPED off a 144 foot cliff. A second later, another stone is THROWN off that cliff. The two hit the water at the same exact instant. What is the initial velocity of the second stone?


    The answer is 40 ft/second according to my book, but for some reason, I never get that.


    I set 4.9t^2 = v*(t-1) + 4.9(t-1)^2
    I also set 144 = 4.9t^2 to get t = 5.421.

    I plug this back into the first equation, and my answer comes out to 10.908 ft/ sec!

    What am I doing wrong here?
     
  2. jcsd
  3. Sep 21, 2006 #2
    youve got the gravitational acceleration wrong...im pretty sure its not 4.9f/s^2...if its 9.8m/s^2
     
  4. Sep 21, 2006 #3
    its approximately 32.2 f/s^2
     
  5. Sep 21, 2006 #4

    radou

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    Interesting, I get the same answer. From the equation of displacement for the first rock we get t = 5.42 seconds. The time of 'flight' of the second stone must be 4.42 seconds. So, x2(t)=v0*4.42+0.5*9.81*4.42^2=144 implies v0=10.9 ft/second.
     
  6. Sep 21, 2006 #5

    OVB

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    OHHH darn I forgot to convert the units. Silly me :P
     
  7. Sep 22, 2006 #6

    radou

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    The one and only unit for length is [m] is a perfect world. :smile:
     
  8. Sep 22, 2006 #7

    HallsofIvy

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    Yes, the acceleration due to gravity is 9.8 m/s2 or 32 ft/s2. QuantumKing, remember that the coefficient of t2 in the distance formula is g/2! 4.9t2 would be correct in the metric system, 16t2 is correct in the imperial system.
     
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