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saifoo.khan

  • 4 years ago

Anyone online?

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  1. saifoo.khan
    • 4 years ago
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    Q4

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  2. shadowfiend
    • 4 years ago
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    Let me sketch out a FBD for you really quick.

  3. shadowfiend
    • 4 years ago
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    Derp heh. A diagram, not a free body diagram. No forces here :p

  4. saifoo.khan
    • 4 years ago
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    It is dealing with v = u + at first, then s = ut + 0.5at^2

  5. shadowfiend
    • 4 years ago
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    |dw:1327119231882:dw| First move is finding when A is at its apex.

  6. saifoo.khan
    • 4 years ago
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    i got 0.5 secs

  7. shadowfiend
    • 4 years ago
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    You can find that by solving \(v = v_0 + at\) for t, and plugging in \(v_0 = 5\) and \(v = 0\) (since velocity is 0 at the apex), and \(-9.8ms^{-2}\) for a, since we are accelerating against gravity.

  8. shadowfiend
    • 4 years ago
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    If you got 0.5s, I assume you're using \(-10ms^{-2}) as an approximate for gravity?

  9. saifoo.khan
    • 4 years ago
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    Check this out..

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  10. shadowfiend
    • 4 years ago
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    I get 2.775 (= 2.8) for the s of B...

  11. shadowfiend
    • 4 years ago
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    You seem to be using 7 for its velocity, it has a v_0 of 8.

  12. saifoo.khan
    • 4 years ago
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    OMG! im blind.

  13. shadowfiend
    • 4 years ago
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    :) It happens. The other s is right, so I'm guessing everything else will work out :)

  14. saifoo.khan
    • 4 years ago
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    i get correct answer now, w.r.t marking scheme! :D 1.5N.

  15. shadowfiend
    • 4 years ago
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    !! Awesome :)

  16. saifoo.khan
    • 4 years ago
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    Yes!

  17. shadowfiend
    • 4 years ago
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    Just watch out for those numbers :) They're sneaky little things.

  18. saifoo.khan
    • 4 years ago
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    Thanks!!! ^_^

  19. shadowfiend
    • 4 years ago
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    No problem!

  20. saifoo.khan
    • 4 years ago
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    what about the (ii) part?

  21. shadowfiend
    • 4 years ago
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    Whoops, let me have a look-see.

  22. mathmate
    • 4 years ago
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    Is it part ii you want to see?

  23. saifoo.khan
    • 4 years ago
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    Yes.

  24. shadowfiend
    • 4 years ago
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    I'll let mathmate help heh. I got distracted trying to track down a song that's been stuck in my head all day >_<

  25. mathmate
    • 4 years ago
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    I have 8t-(g/2)t^2-(5t-(g/2)t^2) = 0.9 which solves for 3t=0.9, or t=0.3 (the g... terms cancel out). Then height of A is 5(0.3)-(g/2)(0.3)^2 =1.0586 (using g=9.81)

  26. saifoo.khan
    • 4 years ago
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    OMG! that's correct. :D

  27. saifoo.khan
    • 4 years ago
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    i was confused about which one to make variable to.. hehe it was "t"/

  28. mathmate
    • 4 years ago
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    Great, I guess I got too many digits, but it's easier to cut than to add! :)

  29. saifoo.khan
    • 4 years ago
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    right.

  30. saifoo.khan
    • 4 years ago
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    thanks!

  31. mathmate
    • 4 years ago
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    Good night then... till next time!

  32. mathmate
    • 4 years ago
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    You're welcome! :)

  33. saifoo.khan
    • 4 years ago
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    :]

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