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Modeling Projection Motion Problem - Calc 3

  1. Jul 19, 2012 #1
    1. The problem statement, all variables and given/known data
    A person is standing 80 ft from a tall cliff. She throws a rock at 80 ft/sec at an angle of 45 degrees from the horizontal. Neglecting air resistance and discounting the height of the person, how far up the cliff does it hit?

    2. Relevant equations
    r = ((xo + vo*cos(~))t) i + (yo + (vo*sin(~)t - ((1/2)gt^2))j
    r = the distance ; xo = initial distance from horizontal direction ; yo = initial distance from vertical direction ; vo = initial velocity; t = time; g = gravity (9.8m/s); ~ = degrees

    3. The attempt at a solution

    The correct answer is 48 feet. I did not get that answer from plugging in all the values in the equation.
     
  2. jcsd
  3. Jul 19, 2012 #2
    What exactly did you plug in?

    Your listed gravity was in m/s, although all given measurments were in feet. Is it possible that the numbers weren't converted?
    Also, your equations have time in them, and you never mentioned how you solved for time.
     
  4. Jul 19, 2012 #3
    Man that was it! I didn't convert the m/s into ft/s!

    For time, I used the first part of the equation r = xo + (vo*cos~)t => 80/(80*cos45) = t = 1.414
    Then I just plugged this into the second equation r = yo + (vo*sin~)t-(.5)(g)(t^2) => r = (80*sin45)(1.414)-(4.9)(1.414)^2) = 70.2ft is the answer i gotten.

    If I replace 4.9 with the converted ft (32.1522) I would get ~48ft!

    Thanks bud!
     
  5. Jul 19, 2012 #4
    The problem can be solved algebraically. Just get rid of t, express vertical displacement in terms of v,g, vertical distance, and angle. A little extra pain is converting dead King's finger unit to SI back and forth. And if you are using calculator to find sin(), cos(), and you were entering 45, make sure it's in degree mode.
     
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