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Projectile launched over cliff

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A projectile is launched with a speed of 50.0 m/s and an angle of 40.0o above the horizontal, from the top of a 75.0 m high cliff onto a flat valley floor at the base of the cliff. Assume that g = 10.0 m/s2 and ignore aerodynamic effects.

    How long does it take for the projectile to reach the valley floor?

    2. Relevant equations

    3. The attempt at a solution

    Not sure on this one
  2. jcsd
  3. Jan 23, 2009 #2
    Hi Jordash, well this problem again is utilities the equations that I gave you in your thread on a particle falling:
    Now you may not be familiar with the fact that if you give something a velocity at an angle, you can break it up in to its horizontal and vertical components using trigonometry. Once you have done that you consider the vertical and horizontal motion of the particle seperatly.

    You should also define which direction you want to be positive, for example is down negative which would would make the acceleration due to gravity negative, it doesn't matter which way you chose, as long as you are consistent :D

    With that try and have an attempt at the question, then post back here with as far as you can get yourself :D
  4. Jan 24, 2009 #3
    Ok, the angle still throws me a little bit off, because wont the displacement (s) be more than 75 meters? Because it is shot up 40 degrees into the air? How do you take that into consideration?

  5. Jan 25, 2009 #4
    Oh k sure. so lets consider it like this, lets set the zero point of vertical displacement at the top of our cliff. Let us also say that the upward direction is positive. Now assuming you are comfortable with breaking the initial velocity in to its components using trigonometry, you will find that the initial vertical component of velocity to be positive (pointing in the upward direction).

    Now we have set the zero point to the top of the cliff, what this means is that the final displacement (which is a vector quantity so must have a direction, in this case up or down) is negative 75m, can you see that. It actually doesn't matter about the hight that it goes up to, the equation does need to know that :D.

    Now (this will give you a little bit of a hint to the correct equation to use) if you chose the correct equation you will actually end up with a positive and negative value for time. obviously it should seem logical that you want the positive value for time, and if you would like I will explain where the other value comes from.

    So try setting up you equation now, paying special attention to the direction in which you values are pointing (eg gravity is acting down so will that be positive or negative in the equation)
  6. Jan 27, 2009 #5
    Sorry i'm still a bit confused I first tried solving for Velocity Final LaTeX Code: v_f = v_i + at and did vf=0+2(-10m/s)(-75m) which gave 1500 and then I tried to solve for time using LaTeX Code: s = \\frac{(v_f + v_i)}{2}t and that came out with .1 second but I don't think I'm doing it right. :(
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