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Rock Launching from a catapult - 2D Kinematics

  1. Sep 27, 2007 #1
    1. The problem statement, all variables and given/known data

    A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H of 33.0 m above sea level, directed at an angle θ above the horizontal with an unknown speed v0

    [​IMG]

    The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 141.0 m. Assuming that air friction can be neglected, calculate the value of the angle θ.
    Calculate the speed at which the rock is launched.
    To what height above sea level does the rock rise?

    2. Relevant equations

    v = v0 +at
    x-x0+ v0t + 1/2at^2
    v^2 = v0^2 + 2a(x-x0)
    x-x0 = 1/2(v0+v)t


    3. The attempt at a solution

    To be honest I really had no idea how to do this one, all I know is that you have to find theta first before you can do anything else I couldn't find that, so I couldn't get anything else. Help would be appreciated.
     
  2. jcsd
  3. Sep 27, 2007 #2
    Well first of all, this is a projectile motion problem. Remember that in such problems, you can resolve the motion into x and y components.

    If you need a little bit of refresher, first read this for conceptual understanding

    http://www.physicsclassroom.com/Class/vectors/U3L2a.html

    Then this for quantitative analysis with java applet

    http://www.ngsir.netfirms.com/englishhtm/ThrowABall.htm

    At any rate, the equations you should use

    [tex]v_x = v_0\cos(\theta)[/tex]
    [tex]v_y = v_0\sin(\theta) - gt[/tex]
    [tex]s_x = s_x_0 + v_xt[/tex]
    [tex]s_y = s_y_0 + v_yt - 0.5gt^2[/tex]

    You have two unknowns to calculate for : the initial velocity, the angle which the rock was launched at and the maximum height of the rock. You can easily find the last part, if you are able to find the initial velocity and the angle, so at the present moment, you have two unknowns.

    You know the projectile remains in flight for 6 seconds. When the projectile lands, you know the y distance is equal to 0. Plug in 0 into the [tex]s_y[/tex] equation, solve for v_y. You now have one equation. Note that you have an initial height.

    Next, you know the projectile travels in horizontal direction of 141.0m during the time of flight. Plug 141.0m into [tex]s_x[/tex] equation, solve for v_x. Note that your initial x distance is 0 in this case.

    You now have two equations and two unknowns. Solve for the initial velocity and the angle.

    Now as for finding the maximum height, if you know calculus, you can derive the [tex]s_y[/tex] equation, set it equal to 0, find the time which the projectile is at the highest time and then plug that time into your [tex]s_y[/tex] equation.

    If not, then you can do this the other way. When does the object reach its maximum height? Throw a rock vertically, and what do you know about the rock when its at its maximum height? Specifically speaking, what is the velocity equal to at that particular point?

    Use the same information for this problem. What is the y-component of velocity equal to when the projectile reaches the maximum height? Plug that particular velocity into your [tex]v_y[/tex] equation, solve for time, then plug that particular time into [tex]s_y[/tex] equation.
     
    Last edited: Sep 27, 2007
  4. Sep 27, 2007 #3
    For clarification is my s initial in the y direction the 33m?
     
  5. Sep 27, 2007 #4
    Correct.
     
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