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Finding Height With Projectile Motion

  1. Sep 18, 2013 #1
    1. The problem statement, all variables and given/known data
    A brick is thrown upward from the top of a
    building at an angle of 10.1◦
    above the horizontal and with an initial speed of 6.74 m/s.
    The acceleration of gravity is 9.8 m/s^2.
    If the brick is in flight for 2.5 s, how tall is
    the building?
    Answer in units of m


    2. Relevant equations
    x=vix*t
    y=viy*t+.5*g*t^2


    3. The attempt at a solution
    y=viy*t+.5*g*t^2
    =(((1/2(9.8m/s^2))2.5s^2)
    =30.625m
    The answer is incorrect, what have I done wrong in my calculations?
     
  2. jcsd
  3. Sep 18, 2013 #2

    lewando

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    A couple things:

    The equation you want to use should be y = yi + viyt + 0.5gt2

    Also, you do not seem to be using viy in the incorrect equation you are using.
     
  4. Sep 19, 2013 #3

    CWatters

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    What lewando said and be careful with the signs.
     
  5. Sep 19, 2013 #4

    HallsofIvy

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    I think you have completely misunderstood the formulas you are trying to use. What became of "viy*t"? You understand that the initial upward velocity is NOT 0 don't you?n What do you think "a brick is thrown upward" means. And you have not used the fact that it is thrown at "an angle of 10.1 degrees above the horizontal". What does that mean and how does it affect your formula?
     
  6. Sep 19, 2013 #5
    Thanks for the reply. But I do not get where yi came from? Can you please explain to me.

    Thanks for the reply. I see that my mistake here is that i assumed viy would automatically equal 0 and that I mixed it up. I also assumed when the problem said the brick is thrown upward it meant being thrown at an angle of 10.1 degrees forming a sort of triangle till it reaches its peak. So what I think it means is that i have to calculate the x distance traveled and the y distance traveled using sin and cos. Am I on the right track?
     
  7. Sep 19, 2013 #6

    lewando

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    I would define:

    yi as the initial height.

    y as the height at time t.



    There are variations of this equation that could lead to confusion.

    d = viyt + 0.5at2, where d = y - yi



    Also, as CWatters pointed out, be careful with signs. The equation I posted:

    y = yi + viyt + 0.5gt2

    as written, uses "g" as -9.81m/s2


    No triangles in projectile motion.

    The x-distance and the x-component of the initial velocity, vix are not important. The y-component of the initial velocity, viy is important. Can you determine what viy is? Yes, a trig function applies.
     
  8. Sep 20, 2013 #7

    CWatters

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    No.

    You use trig to work out the vertical component of the initial velocity and plug that into an equation of motion.

    In this particular problem you can ignore what's happening in the horizontal plane. They only ask for the height of the tower not the horizontal distance travelled.
     
  9. Sep 20, 2013 #8
    Sorry if this sounds like a stupid question but how would I find the y-velocity? And if I do I would have to do
    y-velocity divided by 6.74 m/s times cos10.1?
     
  10. Sep 20, 2013 #9

    lewando

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    The initial velocity is given as 6.74m/s in a direction 10.1° (assumed above the horizontal).

    This velocity vector can be represented as the sum of two vector components: one in the x-axis direction (horizontal) and another in the y-axis direction (vertical).

    Your job is to find the one in the y-axis direction. Here is a good refresher link:
    http://hotmath.com/hotmath_help/topics/components-of-a-vector.html
     
  11. Sep 22, 2013 #10

    CWatters

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    Close. That would be the horizontal velocity not the initial vertical velocity.
     
  12. Sep 23, 2013 #11
    Ok thanks guys, I was finally able to solve it! Thanks for your time guys!
     
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