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Does anyone want to check this? (Stone from David's sling hits Goliath)

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data

    David releases a stone from his sling at an angle of 30 degrees above the horizontal. The length of the sling is 0.88 m and it was taking 0.2s per revolution. If Goliath's eye was (for some reason) at the exact same height as the release point, then how far away was he if he is hit by the projectile?


    2. Relevant equations
    Since the stone hit Goliath's eye the same height it was released at, we are only concerned about the range

    I used the equation range=v^2*sin(2θ) /g

    To get the velocity I first found the circumference, C=2r, with r=0.88m
    (Apparently the stone makes one revolution that equals the circumference of a circle. Which I'm still a little confused about ???)
    Once I found the circumference I divided it by the time, 0.2 s, to get the velocity.

    So, C=2π*0.88m=5.53 meters
    v = C/t = 5.53m/0.2s =27.646 m/s (I only knew to do this because its the only way to get a velocity)

    3. The attempt at a solution
    Finally, Range= (27.646)^2*sin(30) /9.8 =67.54 meters
     
  2. jcsd
  3. May 26, 2015 #2

    nrqed

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    Watch out, you have to take the sin of twice the initial angle theta, so in your case you should have sin(60).
    The rest looks right.
    (by the way, when an object is moving in uniform circular motion (circular at constant speed), then yes, the speed is simply distance/time = circumference/period of rotation)
     
  4. May 26, 2015 #3
    Thank you I used 60 when I did the calculation.
     
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