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Homework Help: I need help with a problem dealing with two dimensional projectile motion.

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data
    In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower height is h = 10.80 m, and that the bulls-eye's horizontal distance is d = 3.4 m from the launch point. (Neglect air resistance.)

    If a jack-o-lantern is given an initial horizontal speed of 3.3 m/s, what are the direction and magnitude of its velocity at the following moments in time? (Let the +x axis point to the right.)

    (a) 0.75 s after launch

    _____________.° (below the +x-axis)

    ____________ m/s

    (b) just before it lands
    _______________ ° (below the +x axis)
    ______________ m/s


    2. Relevant equations
    I am not entirely sure but I believe the relevant equations are....
    Vot + .5gt^2
    d = 3.4m
    h= 10.80m
    It would be great if someone could give me the required equations and method to solving this kind of problem. Thanks.

    3. The attempt at a solution

    I am confused how to proceed or all that is required in the solution of this problem.
    What I have done so far is substituted into the Vot + .5gt^2 to find vertical and horizontal (x and y) distances. I did this by using the equation given .75 seconds accounting for the initial velocity on the horizontal (3.3 m/s). I am unsure how to proceed though or if i am even doing this right. Very confused.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 17, 2010 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The equations you need are:

    (1) [tex]x = v_{0x}t[/tex] and

    (2) [tex]y = v_{0y} - \frac{1}{2}gt^2[/tex]

    In terms of t, (1) can be written:

    (3) [tex]t = x/v_{0x}[/tex]

    If the initial motion is only horizontal, v0y=0, so (2) can be written in terms of time, t:

    (4) [tex]t = \sqrt{-2y/g}[/tex]

    AM
     
  4. Apr 17, 2010 #3
    First, start by drawing a picture of the scenario. The projectile is at a height h above the origin, has initial velocity vector [tex]v_{0x}[/tex] in the x-direction, and weight vector mg in the -y-direction. Also, the target is distance d along the x-axis. Start by breaking any forces on the projectile into x and y components. The only force acting on the projectile is the gravitational force.

    x component
    [tex]\sum[/tex][tex]F_{x}[/tex] = 0 = m[tex]a_{x}[/tex]

    [tex]a_{x}[/tex] = 0 (1)

    [tex]v_{x}[/tex] = [tex]\int[/tex][tex]a_{x}[/tex]dt = [tex]v_{0x}[/tex] (2)

    x = [tex]\int[/tex][tex]v_{x}[/tex]dt = [tex]v_{0x}[/tex]t + [tex]x_{0}[/tex] (3) (where [tex]x_{0}[/tex] = 0)

    y component
    [tex]\sum[/tex][tex]F_{y}[/tex] = -mg = m[tex]a_{y}[/tex]

    [tex]a_{y}[/tex] = -g (4)

    [tex]v_{y}[/tex] = [tex]\int[/tex][tex]a_{y}[/tex]dt = -gt + [tex]v_{0y}[/tex] (5)

    y = [tex]\int[/tex][tex]v_{y}[/tex]dt = -(1/2)[tex]gt^{2}[/tex] + [tex]v_{0y}[/tex]t + [tex]y_{0}[/tex] (6) (where y[tex]_{0}[/tex] = h)

    We need to find the initial vertical speed [tex]v_{0y}[/tex] required to hit the target. We know that when the projectile hits the target, x = 3.4 m and y = 0 m. We can start with equation (3) to find the time of impact when x = 3.4 (t = [tex]\frac{d}{v_{0x}}[/tex]). At t = [tex]\frac{d}{v_{0x}}[/tex], y = 0, so we can solve equation (6) for the initial vertical speed [tex]v_{0y}[/tex]. We can now use equations (2) and (5) to find the x and y components of velocities at any time. The angle of the velocity vector will be negative wrt the +x-axis (since the projectile is falling) and is given by [tex]\theta[/tex] = [tex]tan^{-1}([/tex][tex]\frac{v_{y}}{v_{x}}[/tex]). You can see this by drawing the x and y components of velocity on the diagram, [tex]tan \theta = \frac{v_{y}}{v_{x}}[/tex]. The magnitude of the velocity vector will be the vector sum of the x and y components v = [tex]\sqrt{v_{y}^2 + v_{x}^2}[/tex]
     
    Last edited: Apr 17, 2010
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