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Projectile calculation help

  1. Oct 5, 2004 #1
    Need help with this one:

    A Projectile is launched at 35.0 degrees above the horizontal with an intial velocity of 120m/s. What is the projectiles speed 3.00 seconds later?

    Ive got the x and y compontents...wut now?
  2. jcsd
  3. Oct 5, 2004 #2
    This is what you have:
    [tex]\vec v=v_xi+v_yj[/tex]

    Think of the above like a right triangle with the speed as the hypotenuse. How would you go about finding the length if this were a trig or algebra class?
  4. Oct 5, 2004 #3
    The projectial's horizontal speed would be the x component
    Find the vertial speed by using

    SpeedFinal=solve for this
  5. Oct 5, 2004 #4
    well i tried using Vf = Vi + at ...which has all i need but it doesnt give the correct answer (106m/s)
  6. Oct 5, 2004 #5
    ohh, nevermind i get it...its the resultant of the 2 components :biggrin: thx
  7. Oct 5, 2004 #6
    It wont give you the correct answer. The speed of the projectile is the magnitude of the velocity vector. Velocity vector has two components--velocity in the x direction and velocity in the y direction. You said you found the x and y components already so you have your vector components.
  8. Oct 5, 2004 #7
    Good. You figured it out before I finished typing.
  9. Oct 5, 2004 #8
    lol yea thx! One other thing concerning projectiles...how do you solve for maximum height with only an initial velolcity at a certain angle?
  10. Oct 5, 2004 #9


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    At Max Height, Vy will be 0, you know the vector is tangent to the parabole, if you know your calculus, you know the vector will be horizontal at max height.
  11. Oct 5, 2004 #10
    hmm...still not gettin it
  12. Oct 5, 2004 #11


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    Show your work.
  13. Oct 5, 2004 #12
    Y Components:
    Vi = 7.89
    Vf = 0
    a = -9.8
    t = 1.62
    d = ?

    d = Vit + .5(a)t^2
    0 = 7.89t + .5(-9.8)t^2
    t = 1.62

    d = 7.89(1.62) + .5(-9.8)(1.62)^2

  14. Oct 5, 2004 #13
    lol nevermind im dumb...i got it
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