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Projectile motion

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired at an angle of 60 degrees above the horizontal with an initial speed of 300 m/s. Calculate A) the horizontal distance traveled and B) the vertical height attained in the first 6 s.

    2. Relevant equations

    3. The attempt at a solution
    I did 300cos60=150 so Vi(x)=150 and 300sin60=259.8 so Vi(y)=259.8 but after that it seems like i don't have enough information to continue.
  2. jcsd
  3. Oct 2, 2008 #2
    you need to use constant acceleration formulae

    we know:
    Ux = 150ms-1
    Uy = 259.8ms-1
    a = -9.81ms-2

    First you can calculate how long it's in the air for, we know that its trajectory is symmetrical about the peak, and at the peak it's y velocity is 0ms-1.

    [tex] v = u + at [/tex]

    [tex] t = \frac{v-u}{a} [/tex]

    = [tex] \frac{0 - 259.8}{-9.81} = 26.48s [/tex]

    For it to hit the ground again, it must have been in the air twice that time = 53s.

    now you know for the x plane, a, t and u
    find s (displacement)

    [tex] s = ut + \frac{1}{2}at^2 [/tex]

    hint: half of that equation dissapears when you sub in the values.

    You can do the rest yourself!
    Last edited: Oct 2, 2008
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