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Ball thrown to a cliff

  1. Sep 21, 2009 #1
    A man standing 31 m from the base of a vertical cliff throws a ball with a speed of 38 m/s aimed directly at a point 5 m above the base of the cliff.

    Question:

    How long does it take the ball to reach the cliff? 0.82s

    Neglecting air resistance and the height of the man, calculate the height above the base of the cliff at which the ball hits. 1.66m

    How fast is the ball moving when it reaches the cliff? 37.3m/s

    Here is where I am having trouble...

    At what time does the ball reach its largest vertical height?

    I am sure that I have to find the intitial vertial velocity:

    31 = 38cos(x)(0.82) where x is the angle x = 5.8 degrees


    vy0 = 38sin(x) vy0 calculated to be 3.84

    vf = v0 + at
    0 = 3.84 + (-9.8)t = 0.39s

    time to max height i figured to be 0.39s

    This is the incorrect answer, what did I do wrong? Any suggestions?

    Thanks!
     
  2. jcsd
  3. Sep 21, 2009 #2
    I don't know how you got your value for x but I found a different value by putting x=arctan(A/B) where A = distance between man and base of cliff, and B = height of the cliff at which man is aiming.
     
  4. Sep 22, 2009 #3
    Hey I have a very similar problem. How did you get a time of 0.82s. My numbers were 28m, 7m, and V=33 m/s. I tried using inverse tan 7/28 to get theta, and then used that theta in t = (33 cos theta)/28. but that's wrong, can someone help?
     
  5. Sep 23, 2009 #4
    [tex]v = \frac{d}{t}[/tex]
    therefore
    [tex]t = \frac{d}{v}[/tex]

    Instead, you wrote t=v/d. You multiplied cos(theta) by 33 and divided by 28, but 28 is distance and 33 is velocity. Alternatively, you can get the result by finding the length of the line between the ball thrower and the point on the cliff he is aiming using Pythagorean theorem. (which is basically what you're finding when multiplying 28 by cos(theta))
     
    Last edited: Sep 23, 2009
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