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Thread: Projectile Motion :D

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    Cadet Smilebuffalo's Avatar
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    Projectile Motion :D

    A horizontal drainpipe 6m above sea level empties stormwater into the sea. If the water comes out horizontally and reaches the sea 2m out from the pipe, find the initial velocity of the water, correct to 1 decimal place. let g = 10ms^-2 and neglect air resistance.

    Answer = 1.8ms^-1

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    Executive Member untouchablecuz's Avatar
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    Re: Projectile Motion :D

    the angle is 90 because it is launched horizontally.

    enjoy

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    cbff... Michaelmoo's Avatar
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    Re: Projectile Motion :D

    Quote Originally Posted by untouchablecuz View Post
    the angle is 90 because it is launched horizontally.

    enjoy
    Um... doesn't horizontally mean the angle is 0?

    and x(dot) should be vcos0? y(dot) should be vsin0? (when t = 0)
    Last edited by Michaelmoo; 27 Aug 2009 at 4:12 PM.

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    Prophet 9 FTW clintmyster's Avatar
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    Re: Projectile Motion :D

    Quote Originally Posted by Michaelmoo View Post
    Um... doesn't horizontally mean the angle is 0?
    yeah thats right, not 90
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    cbff... Michaelmoo's Avatar
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    Re: Projectile Motion :D

    Quote Originally Posted by clintmyster View Post
    yeah thats right, not 90
    Wait maybe he takes his angle relative to the vertical. Bit wierd though...

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    Executive Member untouchablecuz's Avatar
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    Re: Projectile Motion :D

    Quote Originally Posted by Michaelmoo View Post
    Um... doesn't horizontally mean the angle is 0?

    and x(dot) should be vcos0? y(dot) should be vsin0? (when t = 0)
    for these Q's, it take it from the vertical

    and notice that y(dot) = vsin0 = vcos(90-0) = vcos90 and the same with x(dot)

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    Cadet Smilebuffalo's Avatar
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    Re: Projectile Motion :D

    so should i be using the angle of projection = 0 or 90? :S

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    Member Prashant Sallan's Avatar
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    Re: Projectile Motion :D

    Quote Originally Posted by Smilebuffalo View Post
    so should i be using the angle of projection = 0 or 90? :S
    works with both, bit what would be the angle if something is 'dropped' vertically down from a height?
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    Executive Member untouchablecuz's Avatar
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    Re: Projectile Motion :D

    no angle, all forces are already in the same vertical plane

    there is only a gravitational force down (ignoring air resistance)

    i.e F=mg => y(double dot) = g

    If wind resistance is taken into account,

    y(double dot) =g-R/m

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    Cadet Smilebuffalo's Avatar
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    Re: Projectile Motion :D

    can someone please show me the entire method of solving this question using the angle = 0.

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    Re: Projectile Motion :D

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    cbff... Michaelmoo's Avatar
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    Re: Projectile Motion :D

    Quote Originally Posted by untouchablecuz View Post
    for these Q's, it take it from the vertical

    and notice that y(dot) = vsin0 = vcos(90-0) = vcos90 and the same with x(dot)
    Just curious. Any particular reaason you take it relative to the vertical?

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    Executive Member untouchablecuz's Avatar
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    Re: Projectile Motion :D

    Quote Originally Posted by Michaelmoo View Post
    Just curious. Any particular reaason you take it relative to the vertical?
    no patricular reason, it seems more logical

    i've taught my self >90% of the 3 unit course and this has always been the way i've done it

    tbh, i always thought everyone did it like me

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    Re: Projectile Motion :D

    mate, when t=0,y=6, so y does not equal to gt^2/2, also you forgot to +c

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