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A ball rolls on a horizontal table and drops. it hits the 0.

  1. Aug 10, 2016 #1
    1. The problem statement, all variables and given/known data
    it hits the ground 0.5s after dropping and at a 40cm distance from the table.
    what is the table height?
    what is the ball initial speed?
    in what speed (with direction) did it hit the ground

    2. Relevant equations


    3. The attempt at a solution
    solved thee first question stuck on the other two
    thx for any help!

    d = v0t + ½gt2

    v0 = 0

    t = 0.5s

    g = 10m/s2

    d = 0 + ½• (10m/s2 • (0.5s)2 = 1.25m

     
  2. jcsd
  3. Aug 10, 2016 #2

    BvU

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    Hi jazz,:welcome:

    You did well on part 1. What equation describes the horizontal motion during these 0.5 seconds ?
     
  4. Aug 10, 2016 #3
    hi ty for your reply!
    I solved question 2 also here's what I did:

    X(t) = X0 + V0xt

    X(t) = 40cm = 0.4m

    0.4m = 0 + V0x • 0.5s

    0.5sV0x = 0.4m

    V0 = 0.8m/s

    they also ask in what direction, what does that mean?
    still struggling on question 3 would appreciate any guidance, thx again!
     
  5. Aug 10, 2016 #4

    jbriggs444

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    The ball will hit the ground at some angle above the horizontal.
     
  6. Aug 10, 2016 #5

    BvU

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    Good. So the ball has 0.8 m/s horizontal speed. How much vertical speed does it have when it hits the ground ?
     
  7. Aug 10, 2016 #6
    where? table.png
     
  8. Aug 10, 2016 #7

    BvU

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    Good thing you made a drawing. If the floor is at the end of the red line, there !
     
  9. Aug 10, 2016 #8
    Vy(t) = V0y – gt

    0 = V0y – 10m/s2 • 0.5s

    V0y = 5m/s

    is that right?
     
  10. Aug 10, 2016 #9

    BvU

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    I would say that ##\ v_y(0) = 0 \ ## as in your first post and therefore ##v_y(0.5) = -5 ## m/s. The minus sign indicates it is going down.

    So you now have horizontal and vertical speeds ##\vec v(0.5) = (0.8, -5) ## m/s. Can you calculate the magnitude of the total velocity and the angle with the horizontal ?
     
  11. Aug 10, 2016 #10
    think I made it:


    Tanα = 5/0.8 → α = 80.9°

    Y = 5sin80.9° = 4.937

    X = 0.8cos80.9° = 0.126

    4.937 + 0.126 = 5.06m/s
     
    Last edited: Aug 10, 2016
  12. Aug 10, 2016 #11

    BvU

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    I would say ##-5/0.8##.

    And ##v_x = 0.8, \ \ v_y = -5 ## which with Pythagoras, gives ##v^2 = 25 + 0.64 \ \Rightarrow \ v = 5.06 ## m/s

    (by the way, 5sin80.9° = 3.28 and 0.8cos80.9° = 0.57 !)
     
  13. Aug 10, 2016 #12
    haha you make my efforts look so complicated when there's a clean simple answer
    and I calculated when calc is on degree, thats how my teacher wants it, I thought I cant use Pythagoras I cant remeber why now, very new to all of this, is what I did legit at all? cause if its is (-) then result would be different
     
  14. Aug 10, 2016 #13

    BvU

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    What I meant to say is that 5 * sin 80.9° = 4.94 and 0.8 * cos 80.9° = 0.79 (o:)).

    And what you meant to say is that 5.063 * sin 80.9° = 5.0 and 5.063 * cos 80.9° = 0.8

    That make us even :smile: ?
     
  15. Aug 10, 2016 #14
    haha, too much thinking made me a bit slow I guess, ty for your help! unfortunately I'm sure Ill need it again lol
     
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