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M1 - Projectiles question Watch

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    (Original post by Daniel__)
    After which you do what exactly?
    rearrange to get u=
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    At somewhat of a loss at this point, anyone mind giving me a hint?
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    (Original post by Daniel__)
    To 16tan35 - 4.9(16/u2cos235) - 9 = 0 ?
    So, 4.9(16/u2cos235)= 16tan35-9

    then 1/u^2=\frac{\cos^235}{4.9\times 16}(16\tan 35-9)

    Work out the RHS, flip it and take the square root.
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    Confused at several parts:

    When t is subbed giving you
    Does the Usin35(16/ucos35) , cancel down to 16Utan35 or 16tan35

    And as for the 4.9(16/u2cos235)= 16tan35-9, not quite sure how you rearranged to get . Forgive me for my terrible mathematical skills.
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    You have a(\frac{b}{cd}) = e

    Divide by a

    Divide by b

    Multiply by d

    and you have 1/c

    reciprocal to get c =
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    (Original post by Daniel__)
    Confused at several parts:

    When t is subbed giving you
    Does the Usin35(16/ucos35) , cancel down to 16Utan35 or 16tan35
    You've got one u in the numerator and one in the denominator, hence they cancel and you're left with the latter.

    And as for the 4.9(16/u2cos235)= 16tan35-9, not quite sure how you rearranged to get . Forgive me for my terrible mathematical skills.
    Note that I did it in two stages.

    Firstly to get 4.9(16/(u^2\cos^2 35))= 16\tan 35-9

    Then I multiplied each side by \cos^235 and divided each side by 4.9\times 16
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    Then I multiplied each side by \cos^235 and divided each side by 4.9\times 16
    After which you flip to get the reciprocal of 1/u2, giving you 4.9*16/cos235(16tan35-9)? After which you solve the RHS and square root it?
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    (Original post by Daniel__)
    After which you flip to get the reciprocal of 1/u2, giving you 4.9*16/cos235(16tan35-9)? After which you solve the RHS and square root it?
    Yep, that will work.
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    Strange tried that earlier and still am now getting a different answer.
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    (Original post by Daniel__)
    Strange tried that earlier and still am now getting a different answer.
    I presume the correct one now, as you've not said.
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    (Original post by ghostwalker)
    I presume the correct one now, as you've not said.
    Oh apologies, its not 29.13.

    I basically did (4.9*(16)/cos235)16tan35-9 then sqrted the answer

    getting 16.04
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    (Original post by Daniel__)
    Oh apologies, its not 29.13.

    I basically did (4.9*(16)/cos235)16tan35-9 then sqrted the answer

    getting 16.04
    Just realised that at one stage you had 16^2 in your calculation (correct) and it dropped to 16 (incorrect). I didn't notice and had carried on with that.

    Also 16tan35-9 should be in the denominator after you flipped it. Make sure your brackets are accurate.
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    Aha thanks so much.
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    A cannon ball fired at an angle of 10º has a range, on a horizontal plane, of 1.25km. Ignoring air resistance, find the speed of projection

    ():s=1250m u=ucos10 a=0 t=t
    (→):s=? u=usin10 a=-9.8 t=t

    How would I go about starting this given that I'm not told the vertical displacement nor the vertical u. I assume eventually once I've resolved it in both directions I'll just use Pythagoras to find the speed.
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    (Original post by Daniel__)
    ...
    You have mis-understood the question

    The horizontal range is 1250

    So when the horizontal distance = 1250, then the vertical height = 0
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    Right thank you.
 
 
 
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