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Catapult Angle

  1. Jun 5, 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 32.0 m above sea level, directed at an angle q above the horizontal with an unknown speed v0.
    The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 158.0 m. Assuming that air friction can be neglected, calculate the value of the angle q.

    Calculate the speed at which the rock is launched.

    2. Relevant equations
    X = Xo+ Vot+.5at^2


    V^2= Vo^2 + 2a (X-Xo)

    3. The attempt at a solution

    So i drew a FBD and got:
    Xo= 0
    X= 158 m
    Yo= 32m
    Y= 0
    a= -9.8 m/s^2

    i tried to use the X=Xo... equation but im not sure what to do with theta. In other words, what does theta equal in this equation? i know its going to be Xsin/cos theta, but dont know what to do.

  2. jcsd
  3. Jun 6, 2007 #2
    You know how long the object was in flight. You also know the horizontal distance traveled in that amount of time. Thus, you are essentially given the horizontal component of the velocity. Now, your only task is to determine the initial vertical component of the velocity. Afterwards, vector addition will reveal the initial velocity.

    If you treat the two components (horizontal and vertical) separately, the vertical component can easily be found by using the famous kinematic equations for constant acceleration.

    After you know the initial velocity, use trigonometry to quickly find the launch angle.
  4. Jun 6, 2007 #3
    ok so i got:

    Theta: 4.25×101 deg

    Calculate the speed at which the rock is launched.
    3.57×101 m/s

    To what height above sea level does the rock rise?
    How do i find the height now?
  5. Jun 6, 2007 #4
    Pretend your catapult is on level ground. How high then does the rock reach? Now, hollow out the ground below that point and stick in the ocean 32m below the ground level.
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