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Projectile motion of mini cannon ball

  1. Sep 14, 2006 #1
    I am not too sure whether I am doing this correctly.

    The problem is basically a mini-cannon aimed at a 60 degree angle, hits a target on the ground 0.8m away. I am suppose to find the initial velocity.

    This is what I did:
    Cut the distance in half, assuming the mini cannon-ball will reach its maximum height at that point (and v = 0). Which is 0.4 m. I then use trig, and do the following: tan60 = y/0.4; y = 0.693m

    vf^2 - vi^2 = 2a(xf-xi)
    vf^2 - 0 = 2(-9.8m/sT=^2)(0.693m)
    vf = 3.68 m/s from top of height to ground
    so vi = 3.68 m/s
    So this is the velocity of the y component?

    Then:
    tan 60 = (3.68 m/s) / x
    xtan(60) = 3.68 m/s
    x = 3.68/tan60 = 2.12 m/s
    v(total)^2 = vy^2 + vx^2 = (3.68 m/s)^2 + (2.12 m/s)^2
    v(total) = 4.25 m/s?

    Is that correct? Did I do anything wrong?

    [I also did this, but I don't think I need it.
    vf = vi + at
    0 = 3.68 m/s + (-9.8 m/s^2)(t)
    t = 0.376 s ]
    ------
     
  2. jcsd
  3. Sep 15, 2006 #2
    The formula for the distance of a projectile with no air resistance is

    s = v^2 * sin(2 * theta) /g so v = (s * g / sin(2 * theta)^1/2
    s = (.8 * 9.8 / .867)^1/2 = 3 m/s

    The time for the projectile to reach max height is v * sin(theta) /g
    since vf = vi - gt and vf = 0 at the top of the trajectory
    Multiply this by 2 to get total time of flight and t = 2 * v * sin(theta) / g
    Also, the total distance traveled is s = v * cos(theta) * t
    Substitute the value for t and s = v^2 * 2 * sin(theta) * cos(theta) /g
    Since 2 * sin(theta) * cos(theta) = sin(2 * theta) you get the formula
    for the distance s =v^2 * sin(theta) / g as stated above
     
  4. Sep 15, 2006 #3

    Doc Al

    User Avatar

    Staff: Mentor

    It is certainly true that the cannonball will reach its maximum height at x = 0.4 m. But it is not true that tan60 = y/0.4. The initial angle of the cannonball's trajectory is 60 degrees, but it moves in a parabolic arc not in a straight line.
     
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