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IsTim

  • 4 years ago

A medieval prince trapped in a castle wraps a message around a rock and throws it from the top of the castle with an initial velocity of 12m/s[42 degrees of above the horizontal]. The rock lands just on the far side of the castle's moat, at a level 9.5m below the initial level. Determine the rock's time of flight.

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  1. King
    • 4 years ago
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    hmmm...do u hav the answer..?

  2. IsTim
    • 4 years ago
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    I don't know how to get it, but the answer is 2.4s.

  3. King
    • 4 years ago
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    ok...wait.....

  4. King
    • 4 years ago
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    is a diagram given?

  5. IsTim
    • 4 years ago
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    Yes!

  6. King
    • 4 years ago
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    can u post it......it will be very helpful.....

  7. IsTim
    • 4 years ago
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    |dw:1327933563123:dw| IT's a lil rough

  8. King
    • 4 years ago
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    i am gettin 1.6

  9. IsTim
    • 4 years ago
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    ?

  10. King
    • 4 years ago
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    i got 2.4!!!!!!

  11. King
    • 4 years ago
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    istim u there?

  12. IsTim
    • 4 years ago
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    How?

  13. King
    • 4 years ago
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    see for horizontal the time of flight is 2usintheta/g and fr vertical it is usintheta/g......so we need total T.O.F. so adding we get 3usintheta/g.....now substitute values..... 3*12*sin42/9.8 36*0.6691/9.8 24.0876/9.8 2.4s!

  14. IsTim
    • 4 years ago
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    fr? TOF?

  15. King
    • 4 years ago
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    time of flight[T.O.F]

  16. IsTim
    • 4 years ago
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    and fr?

  17. King
    • 4 years ago
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    for horizontal motion and vertical motion[fr a projectile there always 2 types of motion horizontal and vertical]

  18. King
    • 4 years ago
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    ok...got it?

  19. IsTim
    • 4 years ago
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    LEt me review your statement. It's a bit hard to understand....

  20. IsTim
    • 4 years ago
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    wats u in 2usintheta/g

  21. IsTim
    • 4 years ago
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    And how did you get those formulas>?

  22. King
    • 4 years ago
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    u is initial velocity

  23. King
    • 4 years ago
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    and those formulas can be derived

  24. King
    • 4 years ago
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    but fr now just remember them

  25. IsTim
    • 4 years ago
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    Sorry. What do I derive those formulas from?

  26. King
    • 4 years ago
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    that is super long....just remember them or google it

  27. IsTim
    • 4 years ago
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    What should i be googling?

  28. King
    • 4 years ago
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    the derivation fr the formulas....range,max.height,t.o.f,etc/

  29. IsTim
    • 4 years ago
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    \[n _{1}\sin \theta_{1}=n _{2}\sin \theta_{2}\]

  30. IsTim
    • 4 years ago
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    is it that one?

  31. IsTim
    • 4 years ago
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    hello?

  32. King
    • 4 years ago
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    http://en.wikipedia.org/wiki/Projectile_Motion

  33. King
    • 4 years ago
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    ok?

  34. IsTim
    • 4 years ago
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    I never learnt that formula in class. I'm doing an exam, so what is my best alternative?

  35. King
    • 4 years ago
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    idk....

  36. IsTim
    • 4 years ago
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    Any opinions mashy?

  37. anonymous
    • 4 years ago
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    wait lemme read it again.. i was just reading the comments :D

  38. IsTim
    • 4 years ago
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    k

  39. anonymous
    • 4 years ago
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    nah.. i can't give a better solution sorry.. you need to know the dynamics of a parabolic tragectory!

  40. IsTim
    • 4 years ago
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    Bugger. Ok. I'll look thru my textbook more thoroughly then.

  41. IsTim
    • 4 years ago
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    There isn't an example with that formula in it. Argh

  42. IsTim
    • 4 years ago
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    The initial Velocity is the component of the original velocity right?

  43. IsTim
    • 4 years ago
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    wait, why is it multiplied by 3?

  44. King
    • 4 years ago
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    READ MY COMMENT!

  45. IsTim
    • 4 years ago
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    eh. which one?

  46. anonymous
    • 4 years ago
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    |dw:1327927068621:dw| looking in the vertical direction,we have usin42 and acceleration g downwards at point A,final velocity for path till A becomes o so apply equations for uniform acceleration v-u/t=a the SAME TIME is taken for stone to come to the same height so total time for 1st part is 2*usin42/g THIA GIVES US T1 the rock doesnt stop once it reaches the height from which the prince threw it does it? so we have a 2nd part in this problem |dw:1327927440627:dw| when it reaches point b(from where prince originally threw) at the same level, THE VELOCITY OF ROCK BECOMES SAME AS THE VELOCITY WITH WHICH PRINCE THREW that is usin42 now again apply equations for uniform acceleration |dw:1327927625023:dw| T1+T2 is the total time taken by rock! u have any doubt u may ask

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