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Simple 2D Motion, rock problem

  1. Oct 5, 2013 #1

    Radarithm

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    1. The problem statement, all variables and given/known data

    A stone is thrown horizontally from a height h above the ground (h > 0). The stone hits the ground after having traveled a horizontal distance x (x > 0). What is the expression for the initial speed v of the stone? Neglect air resistance. (please use the variables x and h and the constant g in your answer)

    2. Relevant equations

    h = h0+v0y sin([itex]\theta[/itex])×t+½at2
    x = x0+v0x cos([itex]\theta[/itex])×t+½at2
    v2x = v20x + 2a(x - x0)
    v2h = v20h + 2a(x - x0)

    3. The attempt at a solution

    The velocity when x = 0 (after the rock hits the ground) is 0, so I used that as v in the 3rd equation. a is simply -g (gravity). After simplifying these two I got: v0x = [itex]\sqrt{2gx}[/itex] and v0h = [itex]\sqrt{2gx}[/itex]

    I am obviously doing something wrong; I can't find a way to relate h, x , and g to find v0. Any help would be appreciated.

    edit: x0 is equal to 0.
     
  2. jcsd
  3. Oct 5, 2013 #2

    SteamKing

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    Show your work.
     
  4. Oct 5, 2013 #3

    Radarithm

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    v2x = v20x + 2a(x - x0x)
    since v = 0 when the rock hits the ground, 0 = v20x + 2a(x - x0x)
    x0 is also 0, and a is gravity, 9.8 m/s, therefore:
    0 = v20x + 2gx
    v20x = 2gx
    v0x = [itex]\sqrt{2gx}[/itex]

    and the same applies to the equation that deals with h (height)
    I only just woke up so I can't really think straight :frown:
     
  5. Oct 5, 2013 #4
    Why do you think vx is 0 ? Is there a force in x-direction which produces a deceleration such that the x-component of velocity on reaching the ground becomes zero ?
     
    Last edited: Oct 5, 2013
  6. Oct 5, 2013 #5

    Radarithm

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    I obviously need to redo this problem.
     
  7. Oct 5, 2013 #6
    but you need to think in correct direction...

    What are the horizontal and vertical components of initial velocity (Assuming it to be 'v') ?
     
  8. Oct 6, 2013 #7

    Radarithm

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    Well the acceleration decreases in the y direction, causing the rock to hit the ground; but that's still not answering the question; I'll skip this one and come back to it later.
     
  9. Oct 6, 2013 #8
    The acceleration remains constant in the vertical direction.It is 'g' i.e 9.8m/s2.

    The velocity of the rock when it hits the ground is not zero .The horizontal component of velocity does not change during the flight.It is the vertical component of velocity which changes due to acceleration 'g'.
     
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