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Projectile motion and trajectory

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data

    A car is drove off a cliff at 8 m/s. it lands in the sea 3 seconds later.

    2. Relevant equations

    a) how far does it land?
    b)how high is the cliff? (g = 10m/s^2)
    c) at what angle relative to the horizontal does it land?

    3. The attempt at a solution
    convention chosen : upwards is positive
    initial velocity= 0m/s
    a= -g
    Vx=8m/s

    a) FROM HORIZONTAL COMPONENT

    Vx= x(displacement)/ t( time)
    8m/s = x/3s
    x=24m

    b) FROM VERTICAL COMPONENT
    using the eq'n s=ut + 0.5at^2
    ie. y = 0.5 gt^2

    y = 0.5 (-10)(9)
    y= - 45 m

    c) ** my problem is that im not sure which eq'n to use

    since its relative to the horizontal there is only one... i think.

    Vx = Vt cosθ
    i have t= 3s
    i have x=24m
    i dont have V ,or at least i dont know what V exactly is.

    Help please?:confused:
     
    Last edited: Oct 9, 2011
  2. jcsd
  3. Oct 9, 2011 #2

    ehild

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    Homework Helper
    Gold Member

    It will be clear if you make a drawing indicating the initial and final positions of the car. You need the angle of the displacement vector, not the angle of the initial velocity which is horizontal.

    ehild
     
  4. Oct 27, 2011 #3
    The x-component of the velocity is just the initial 8m/s.


    The final velocity of a uniformly accelerated object is
    Vf=Vi+gt=0+(-9.8m/s^2)*3s=-29.4m/s
    which is the y-component.

    Then the angle should be the angle made up from those vector components: tan(8/24), or is it tan(24/8), can't think straight about it at the moment.
     
  5. Oct 28, 2011 #4
    i solved it already :/ a long time ago.
    am i supposed to delete the post then?
     
  6. Oct 28, 2011 #5

    ehild

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    Homework Helper
    Gold Member

    You should have written that the problem was solved.

    ehild
     
    Last edited: Oct 28, 2011
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