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M1 Projectiles

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    If I don't get part A correct, I'm screwed for the rest of the questions.

    Is it tan inverse 6/8 ?? ie 36.9 degrees? (3SF)

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    Where do you get the 8 from? Use toa.
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    (Original post by Hopple)
    Where do you get the 8 from? Use toa.
    So its tan inverse 6/10?

    IDK but I put the velocity of 8 as part of the triangle then assumed 10 must have been the hyptonuse, so i thought 8 was the adjacent :|
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    (Original post by MSI_10)
    So its tan inverse 6/10?
    Yeah, that's it.
  5. Offline

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    yes tanα is 6/10
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    (Original post by Hopple)
    Yeah, that's it.
    Aight thanks.

    Am I starting part b probably; to get time, first need to use S. Do I use 6 or 10..?

    Then SUVAT (S=ut+o.5at^2) use quadratic formula to get t?
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    (Original post by MSI_10)
    Aight thanks.

    Am I starting part b probably; to get time, first need to use S. Do I use 6 or 10..?

    Then SUVAT (S=ut+o.5at^2) use quadratic formula to get t?
    For part b, are you looking at vertical or horizontal motion (or both)? And yes, that suvat equation will do.
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    (Original post by Hopple)
    For part b, are you looking at vertical or horizontal motion (or both)? And yes, that suvat equation will do.
    I did this..
    S=-6
    a=-9.8
    U=8sin31..

    -6=8sin31t-4.9t^2
    -4.9^2+8sin31t+6=0
    got t=1.60 or t=-0.763

    Answer t=1.60..?
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    (Original post by MSI_10)
    I did this..
    S=-6
    a=-9.8
    U=8sin31..

    -6=8sin31t-4.9t^2
    -4.9^2+8sin31t+6=0
    got t=1.60 or t=-0.763

    Answer t=1.60..?
    Make sure you're measuring everything in the same direction.
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    (Original post by Hopple)
    Make sure you're measuring everything in the same direction.
    :\
  11. Offline

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    (Original post by MSI_10)
    :\
    Even though everything's going downwards, you've made displacement and acceleration be in the opposite direction to initial velocity.
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    (Original post by Hopple)
    Even though everything's going downwards, you've made displacement and acceleration be in the opposite direction to initial velocity.
    Oh that's what you meant, yeah i get what you mean now thanks

    So t=0.763
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    Hmm, can anyone check if im doing part c correctly.

    It's asking for distance.. do I use the time obtained from part b, use S=ut+1/2at^2 but use 8cos31 for u instead of 8sin31..?
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    (Original post by MSI_10)
    Hmm, can anyone check if im doing part c correctly.

    It's asking for distance.. do I use the time obtained from part b, use S=ut+1/2at^2 but use 8cos31 for u instead of 8sin31..?
    Yes, but remember there are no forces acting horizontally here.
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    (Original post by finality)
    Yes, but remember there are no forces acting horizontally here.

    Alright so I used 8sin31
    I get the answer as 6m. Something tell's me this aint right; it seems to high :\

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