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Projectile Motion

  1. Jun 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A bullet is fired from a cliff 36m above the ground. If the bullet is fired at an angle of 25 above the horizontal, and has a muzzle velocity of 80m/s, what is the velocity of the bullet as it hits the ground?

    d = 36m
    vi = 80m/s
    angle = 25 degrees

    2. Relevant equations



    3. The attempt at a solution
    Horizontal: 80 cos 25 = 72.5

    Vertical: muzzle: 80 sin 25 = 33.8

    0^2 = 33.8^2+2ad <-- Top of the projectile motion
    0= 1142.44+2(-9.8)d
    d= 1142.44/19.6
    d= 58.28

    height = 58.28+36 = 94.28
    Vf^2= vi^2+2ad <-- Top of the parabola = 0m/s
    vf^2= 2ad
    vf^2= 2(9.8)(94.28)
    Vf^2= 1847.88
    Vf= 42.99
    R^2 = 42.99^2+72.5^2 = 7105.25
    R = 84.3

    I was just wondering if I did it right...

    Thank you very much,
     
  2. jcsd
  3. Jun 20, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me, but you could have calculated the change in vertical speed in one step using the same formula (Vf^2 = Vi^2+2ad). Let a = -9.8 m/s^2 and d = -36 m.

    (You could also have used conservation of energy.)
     
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