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Supposly simple balistic motion prob

  1. Feb 23, 2013 #1
    At what angle relative to the horizon should you throw a ball at V0=15 m/s (from the ground of course) in order for it to hit back the ground a distance of 21m away?

    I found that 2θ=66.16+2∏k, and so θ=33.08+∏k, but the truth is that θ=56.92 is also a correct answer! How could I get this answer from the equations??

    y(t)=0 at hit so: 0=Vosinθt-0.5gt^2 ... t(impact) = 2V0sinθ/g or 0

    back to x: 21=(2sinθcosθVo^2)/g = sin2θ*V0^2 / g and so sin2θ=21g/V0^2
  2. jcsd
  3. Feb 23, 2013 #2
    This follows from sin x = sin (∏ - x).
  4. Feb 23, 2013 #3
    wow you're right... so why do we say sin has a period of 2∏ if we can see that angles that are seperated less than 2∏ have the same value?
  5. Feb 23, 2013 #4
    f(x) is said to be a-periodic if f(x) = f(x + a). For f(x) = sin x, this works when a = 2∏.

    f(x) = f(a - x) is a different property, not periodicity.
  6. Feb 23, 2013 #5
    ok so what is this property? where does it come from?
  7. Feb 23, 2013 #6
    I am not sure whether it has a special name. You could say that it comes from the definition the sine function on the unit circle. Then it is obvious that the ordinate at some angle a equals the ordinate at (∏ - a).

    Another way to see this is by using the identity sin (∏ - a) = sin ∏ cos a - cos ∏ sin a = sin a, because sin ∏ = 0, and cos ∏ = -1.
  8. Feb 23, 2013 #7
    tnx a lot!
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