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Help with projectile physics

  1. Apr 12, 2004 #1
    projectile is fired w/an angle of 28.0degrees above the horizontal and from height 46.0m agove the ground. the projectile strikes ground w/a speed of 1.95xVo. Find Vo

    I started using the eqution
    (Vfy)^2=(Voy)^2+2*Ay*(delta Y)

    Voy=Vosin(theta)=.47Vo
    Vfy=1.954*Vosin(theta)=.92Vo

    (.92Vo)^2=(.47Vo)^2+(2*-9.8*-46)
    Vo=37.9

    but my answer is wrong and I can't think of another way to answer the problem.....Any ideas?

    Thanks for your time!
     
  2. jcsd
  3. Apr 12, 2004 #2
    What is Vo? Is it the velocity of projection, or, is it just a constant?
     
    Last edited: Apr 12, 2004
  4. Apr 12, 2004 #3
    There's no guarantee that the projectile hits the ground at the same angle as it was launched!

    cookiemonster
     
  5. Apr 13, 2004 #4
    However, for this problem, I think we should assume that the projectile hits the ground at the same angle as it was launched....


    Sridhar
     
  6. Apr 13, 2004 #5
    Why should that be assumed?

    cookiemonster
     
  7. Apr 13, 2004 #6
    Not only are you not supposed to make that assumption, it is wrong! The projectile is travelling in a parabola.........

    You need to use the conservation of energy equation:

    [tex]0 = \Delta E_m = \Delta E_p + \Delta E_k[/tex]
    [tex]0 = mg(0 - h) + \frac{1}{2}m(v_f^2 - v_0^2)[/tex]
    [tex]2gh = (1.95v_0)^2 - v_0^2 = 2.8025v_0^2[/tex]

    The initial velocity is 17.94m/s (for g = 9.8m/s2).
     
    Last edited: Apr 13, 2004
  8. Apr 13, 2004 #7
    Thanks for the help!
     
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