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Angle that the stone is thrown off of the cliff?

  1. Jun 4, 2009 #1
    1. The problem statement, all variables and given/known data
    A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 16.0 m/s. The cliff is h = 62.0 m above a flat horizontal beach.

    How long after being released does the stone strike the beach below the cliff?

    With what speed and angle of impact does the stone land?

    How do I solve without knowing the angle that the stone is thrown off of the cliff?
     
  2. jcsd
  3. Jun 4, 2009 #2

    Cyosis

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    The angle is given by the bolded part.
     
  4. Jun 4, 2009 #3
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    so the angle is 0 or 180?
     
  5. Jun 4, 2009 #4

    Cyosis

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    Technically zero, but what do you know about the components if an object is thrown horizontally? You don't need to do any trigonometry.
     
  6. Jun 4, 2009 #5
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    I tried working it without using trig but I got the wrong answers.
     
  7. Jun 4, 2009 #6

    Cyosis

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    Show your steps so we can have a look at it.
     
  8. Jun 4, 2009 #7
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    Ok so you said I didnt need to use trig so to get the time i simply divided 62 by 9.8 and got 6.3 seconds.
     
  9. Jun 4, 2009 #8

    dx

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    And why do you think that would give you the time?
     
  10. Jun 4, 2009 #9

    Cyosis

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    That is not right, 62 meters divided by 9.8 m/s^2 gives 6.3 s^2. Do you know of any kinematic equations that give you displacement as a function of time for a uniformly accelerating object?
     
  11. Jun 4, 2009 #10
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    Because i figured if the height is 62m and the gravitational constant is 9.8m/s dividing would give me the time of the fall.
     
  12. Jun 4, 2009 #11

    Cyosis

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    Well if you had just looked at the dimensions you would have figured it was wrong. As I said your calculation yields 6.3 seconds squared, which is not the unit of time.

    Edit: the gravitational constant is 9.8 m/s^2 not 9.8m/s.
     
  13. Jun 4, 2009 #12

    dx

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    So you simply guessed? 9.8 m/s² is the acceleration due to gravity. Do you recognize this equation: s = ut + ½at² ?
     
  14. Jun 4, 2009 #13
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    yes i recognize that equation
     
  15. Jun 4, 2009 #14

    Cyosis

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    What are s, u and a in this problem?
     
  16. Jun 4, 2009 #15
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    s=62 u=? a=-9.8m/s
    I tried using this equation at first but ran into the same problem because i was unsure of what u is.
     
  17. Jun 4, 2009 #16

    dx

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    That's the equation to use here. This is what it means: if a particle startes out with speed u (in some direction), and has an acceleration a(in that direction), then the distance travelled by the particle after time t (in that direction) is s = ut + ½at².
     
  18. Jun 4, 2009 #17

    Cyosis

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    You should know what u is after reading dx's post. One thing though if you set s=62 then a needs to be positive or if you take a to be negative then s=-62. Also I generally use this notation:

    [itex]
    s=\frac{1}{2}at^2+v_0t+s_0
    [/itex]

    The v_0 symbol is slightly more suggestive than the u symbol.
     
  19. Jun 4, 2009 #18
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    so would u=16?
     
  20. Jun 4, 2009 #19

    Cyosis

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    In which direction is gravity accelerating the stone? In which direction is the stone moving with 16m/s?
     
  21. Jun 4, 2009 #20
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    actually horizontally
     
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