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Projectile motion

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball is kicked at a velocity of 40.0 ms at 60 degrees above the horizontal/

    a) When is the ball 25.0m above the ground?

    3. The attempt at a solution

    First i broke them down into there x and y components but having problems on the (25.0m) part.

    I did Vix = 40 cos 60 = 20
    Viy = 40 sin 60 = 34.6

    First i tried to put it as finding the maximum height

    -Time-

    Y component

    V=Vi + at
    t = V-vi/a

    a= -9.8
    V= 0 < (0.0ms velocity at peak)

    t = 0 - 34.6/-9.8
    t = 3.5s

    So im not sure if 3.5 is the answer (no answer sheets) since that im stating that 25.0m is below the peak.

    I also used the time to figure out the max. height reached which turned out to be 61m so then i went half of 61 = 30.5m and then halved 3.5 < (since it was to the peak) to then equal 1.7s.

    Totally confused at the moment :S

    Thanks
     
  2. jcsd
  3. Oct 31, 2009 #2

    Kurdt

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    What you will need is a kinematic equation that involves the velocity the acceleration the time and the distance. You know the initial vertical component of velocity the acceleration and the height (i.e. distance) you want to reach and so you can then solve for the time.
     
  4. Oct 31, 2009 #3
    Cant seem to find an equation that works well with the values i have. There's always a component which i don't have and can't find without time in the equations i have or vice versa to finding time.
     
  5. Oct 31, 2009 #4

    Kurdt

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    There is an equation you can use. Like I say you're after the time when the ball is 25 metres above the ground so you have the distance. You have the initial velocity for the veritical component and you know the acceleration. So you need an equation with those four things in it so you can solve for time.
     
  6. Oct 31, 2009 #5
    im trying to use y = yo + vit + 0.5at^2 which simplifies to y = vit + 0.5at^2 which then for to i used

    t = 2(y-vi)/a = 2(25-34.6)/-9.8

    t = 1.9s


    Not sure if that is the answer but that's the equation with all of its components in it that i'v found.
     
  7. Oct 31, 2009 #6

    Kurdt

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    That is the correct equation but you will have to solve a quadratic in t.
     
  8. Oct 31, 2009 #7
    quadratic of t? doesn't the equation already cancel out t^2 leaving t as an individual component?
     
  9. Nov 1, 2009 #8

    Kurdt

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    No. And if you think about it you will realise that you need two solutions for time anyway. The first time is when the ball goes up and the second is when it is coming back down.
     
  10. Nov 1, 2009 #9
    ok thanks
     
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