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anonymous

  • one year ago

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes time 0.420 s to pass this window, which is of height 1.90 m. Question- How far is the top of the window below the windowsill from which the flowerpot fell?

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  1. anonymous
    • one year ago
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    @UnkleRhaukus

  2. anonymous
    • one year ago
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    @johnweldon1993

  3. UnkleRhaukus
    • one year ago
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    |dw:1442325952193:dw|

  4. UnkleRhaukus
    • one year ago
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    what are the forces acting on the flower pot during the motion?

  5. anonymous
    • one year ago
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    you mean the gravity?

  6. anonymous
    • one year ago
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    Take the free fall acceleration to be 9.80m/s^2 .

  7. UnkleRhaukus
    • one year ago
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    yeah gravity is the only force acting here!

  8. UnkleRhaukus
    • one year ago
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    |dw:1442326446681:dw|

  9. UnkleRhaukus
    • one year ago
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    by newtons2nd the sum of the forces is equal the mass times its acceleration ∑ F = ma = mg (if we define the y axis pointing downwards ) so a = g

  10. UnkleRhaukus
    • one year ago
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    we get y(t) = y_0 + v_y0 t + 1/2 g t^2

  11. anonymous
    • one year ago
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    hmmm..can you show me how you derived to that equation

  12. anonymous
    • one year ago
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    the one i know is d=Vit + 1/2at^2

  13. UnkleRhaukus
    • one year ago
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    a = y'' = g v = y' = ∫ g dt = gt +c y = ∫ (gt + c) dt = 1/2 gt^2 +c_1t + c_2

  14. UnkleRhaukus
    • one year ago
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    y(0) = y_0 --> c_2 = y_0 v(0) = v_y0 --> c_1 = v_y0

  15. anonymous
    • one year ago
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    ok...how can we solve this ^^ using that formula

  16. UnkleRhaukus
    • one year ago
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    y(t_1) = y_0 + v_y0 (t_1) + 1/2 g (t_1)^2 y(t_2) = y_0 + v_y0 (t_2) + 1/2 g (t_2)^2 D = y(t_2) - y(t_1) = v_y0 (t_2-t_1) + 1/2 g (t_2-t_1)^2 is this like your formula?

  17. anonymous
    • one year ago
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    yup

  18. UnkleRhaukus
    • one year ago
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    so D is the displacement? and (t_2 - t_1) is the time it takes of the pot to fall past the window

  19. anonymous
    • one year ago
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    absolutely

  20. UnkleRhaukus
    • one year ago
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    how much displacement is is this example?

  21. anonymous
    • one year ago
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    1.90m

  22. UnkleRhaukus
    • one year ago
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    so we have ( v_i = v_y0 ) D = v_i t + 1/2 g t^2 lets solve this equation for v_i, the velocity of the pot plant when it was falling past the top the the window and Then, calculate with D = 1.90 m t = 0.420 s g = 9.80 m/s^2

  23. UnkleRhaukus
    • one year ago
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    what do you get when you rearrange the equation ?

  24. anonymous
    • one year ago
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    2.4658

  25. anonymous
    • one year ago
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    m/s

  26. anonymous
    • one year ago
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    is that we only need?

  27. UnkleRhaukus
    • one year ago
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    now we have the initial velocity , we still need to work out the height to the window sill

  28. UnkleRhaukus
    • one year ago
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    Do you have formulas you can just use, or is the derivation helpful?

  29. anonymous
    • one year ago
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    hmmm.. idk which formula to use lol

  30. UnkleRhaukus
    • one year ago
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    we need to find the distance to the windowsill from the lower window, we have the velocities

  31. UnkleRhaukus
    • one year ago
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    do you have a set of 5 equations of projectile motion?

  32. UnkleRhaukus
    • one year ago
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    or i suppose we can say the the pot started on the sill at rest, and was falling at 2.4658 m/s, at the top on the next window down how long was it falling ?

  33. anonymous
    • one year ago
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    we dont learn projectile from now so i really dont know

  34. anonymous
    • one year ago
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    @rishavraj

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