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Kinematics: Free Fall

  1. Oct 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A hockey puck slides off the edge of a table with an initial velocity of 20 m/s. The height of the table above the ground is 2 m. At what angle below the horizontal does the puck hit the ground?

    2. Relevant equations
    d=Vi(t)+(1/2)at^2
    a^2+b^2=c^2

    3. The attempt at a solution
    Horizontal Values: Vi=20Cos(theta), a=0, Xi=0m, Xf=?
    Vertical Values: Vi=20sin(theta), a=-9.81m/s^2, Xi=2m, Xf=0

    1) 2=20sin(theta)t+(0.5)(-9.81)t^2
    2)1/10=sin(theta)t+(-4.905)t^2
    3) 1/10 x 1/(-4.905)=sin(theta)t+t^2

    Stuck after solving that part.
     
  2. jcsd
  3. Oct 15, 2015 #2

    SteamKing

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    The puck is initially 2 m above the ground when it slides off the table. How long does it take for the puck to hit the ground?

    Hint: the amount of time it takes the puck to fall is not affected by its horizontal velocity.
     
  4. Oct 15, 2015 #3
    I thought there is a horizontal velocity? I'm unsure what you mean, if I use the horizontal value then I do not have the displacement for it. I'm attempting to find the time using the vertical initial velocity, but i have two missing variables: theta and t. I can't do a quadratic. I'm unsure if there even is an initial vertical velocity.
     
  5. Oct 15, 2015 #4
    Wait, do I have my initial velocities backwards?
     
  6. Oct 15, 2015 #5

    SteamKing

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    I never said there wasn't a horizontal velocity, because the puck comes flying off the table at 20 m/s.

    What is important to remember is, the puck takes the same amount of time to fall 2 m whether it comes flying off the table at 20 m/s or 200 m/s. The puck is coming off the table horizontally, so there is no initial upward velocity component for gravity to overcome. If there is no initial upward velocity component for gravity to overcome, then what is the puck's initial vertical velocity?

    If you can't "do" a quadratic, there is a handy formula called, surprisingly enough, the "quadratic formula" which can get you over that hump.
     
  7. Oct 15, 2015 #6
    I meant I did not have the values required to do the quadratic, no need for the ignorant comment. There's what I wanted to know, initial vertical velocity is 0m/s. Thank you, I get confused with free fall off a platform and projectiles.
     
  8. Oct 15, 2015 #7
    I'm stuck again, frustrating to be stuck on a simple question like this. Time calculated to be 0.63s

    d=20cos(theta)(0.63)

    I'm missing 2 values, how do i solve it from here on?
     
  9. Oct 15, 2015 #8

    SteamKing

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    It helps to define your terms.

    What is d supposed to be? What you have written is different from what you wrote in the OP.

    Remember, you're supposed to be solving for the angle below the horizontal at which the puck strikes the ground. Is this θ, or something else?
     
  10. Oct 15, 2015 #9
    This is the displacement equation in the horizontal: d=Vi(t)+(1/2)at^2

    Since a=0 in the horizontal in cancels (1/2)at^2. I am trying to solve for θ, but this is seemingly the only equation that fits. I feel as though I'm overcomplicating this.
     
  11. Oct 15, 2015 #10

    SteamKing

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    You can find the horizontal and vertical components of the velocity of the puck as it hits the ground. These should allow you to construct a right triangle using those velocity components as the sides. From there, you should be able to calculate the angle of impact.

    Remember, the angle desired is supposed to be measured below the horizontal, so make sure to measure this angle from the correct reference.
     
  12. Oct 15, 2015 #11
    So, it's the angle of impact, not the angle below the table after initial release? Isn't that the angle above the horizontal though? Sorry to be bothersome, but do you mind explaining how to know which angle of reference is to be found by the wording when doing these types of equations?
     
  13. Oct 15, 2015 #12

    SteamKing

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    You've got to read and re-read what the problem asks for, and not jump to any hasty conclusions.

    The OP asks this:

    "At what angle below the horizontal does the puck hit the ground?"

    Hit the ground ≠ below the table after initial release. That was an assumption you made, which unfortunately was in error.

    In the second part, about the reference from which the angle is measured, you are assuming that 'the horizontal' is the same as a plane which coincides with the ground. You can measure the angle of impact from the ground up, but the problem here specifically asks for the angle measured below the horizontal. The horizontal can be any plane above and parallel to the ground, like the plane of the table top, for instance.
     
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