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Projectile Motion of a Cannonball Problem

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A canon is set at an angle of 45 degrees above the horizontal. A cannonball leaves the muzzle with a speed of 2.2 x 10^2 m/s. Air resistance is negligible. Determine the cannonball's

    A)Maximum height
    B)Time of Flight *** this is the one im stuck on
    C)Horizontal Range (to the same vertical level)
    D)Velocity at Impact

    The answers are: A) 1.2 x 10^3 m
    B) 32 s
    C) 4.9 x 10^3 m
    D) 2.2 x 10^2 m

    2. Relevant equations

    vertical equations
    v2y^2 = v1y^2 +2ay[tex]\DeltaD[/tex]
    Dy=vfy*t - 1/2at^2

    horizontal equations
    vix =dx/t

    3. The attempt at a solution

    A) vi y = (sin 45)(220m/s)
    = 155.56 m/s

    vf^2 = vi^2 +2ad
    **rearrange to solve for d
    vf^2 - vi^2 / 2a= d
    0 - (155.56m/s)^2 / -19.6
    = 1234.64 m
    = 1.2 x 10^3 m

    ** so finding the maximum height isnt the problem ^
    i cant seem to find the time :grumpy:
    1. The problem statement, all variables and given/known data

    vi x = (cos 45)(220m/s) vi y = (sin45)(220m/s)
    = 155.56 m/s = 155.56m/s
    t= ? a = -9.8 m/s/s
    d=? vf y (at peak) = 0 m/s
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 26, 2010 #2
    Please edit your post to remove the huge space.

    You first need to split the diagonal velocity into it's horizontal and vertical components.

    Once you know the vertical speed you can use vf = vi + at to calculate the time to the highest point.

    Where vf = 0m/s (vertical speed at peak is 0), vi = initial vertical speed, a = -9.8m/s^2

    Plug in your values and rearrange to give the time to the peak or maximum height.

    Then using d = 0.5(vi+vf)t, you get the maximum height d metres.

    For the descent phase, using vf^2 = vi^2 + 2ad, you can insert the initial speed vi (0m/s), the acceleration a (+9.8m/s) and height d (m) you can get the final velocity.

    Once you have that, again using vf = vi + at you can get the time of the descent phase.

    Add the two times together and you have the full time of the flight.

    Because you have split the velocity into it's horizontal and vertical components and now you have the time of the flight, you can use d = 0.5(vi+vf)t to get the horizontal distance. (Ignoring air resistance means there is no deceleration in the horizontal component so vi = vf.)

    Again, the velocity in the horizontal direction is constant, so you know that. You also have the final speed from earlier so you just need to combine them using a^2 + b^2 = c^2 to give you the final impact velocity.

    See how you go with that, I haven't checked any values so see how you go.

    Last edited: Sep 26, 2010
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