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Thread: ball height

  1. #1
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    ball height

    the height, h cm, of a ball moving up and down in a harbour is given by h=30+20sintπ
    where t is in seconds and the angle is in radians.
    (i) find an expression for all the positive times at which the height of the ball is 20 m.
    (ii) find the fourth time at which h=20. give your answer correct to three decimal places


    so what i tried is 20=30+20sintπ
    the answer at the back of the book says 1/6 +2(n) and 5/6 +2(n). i am not sure where this comes from. i do not know how to do part(ii)
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  2. #2
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    Re: ball height

    Your answer is not an expression of time. It is a formula using time. You need to solve for $t$ in your expression:
    $20 = 30 + 20 \sin (t\pi)$
    $-10 = 20\sin (t\pi)$
    $\sin (t\pi) = -\dfrac{1}{2}$

    So, for what values of $\theta$ do you have $\sin \theta = -\dfrac{1}{2}$? That occurs in the 3rd and 4th quadrants, and the reference angle is $\dfrac{\pi}{6}$. For the 3rd quadrant angle, it will be $\pi + \dfrac{\pi}{6}$. For the fourth quadrant angle, it will be $2\pi - \dfrac{pi}{6}$. Additionally, these results are periodic with period $2\pi$. So, we have:

    $\theta = \dfrac{7\pi}{6} + 2\pi n = t\pi$ which gives $t = \dfrac{7}{6}+2n$
    Or
    $\theta = \dfrac{11\pi}{6} + 2\pi n = t\pi$ which gives $t = \dfrac{11}{6} + 2n$

    So, the book's answers are incorrect.

    As for the fourth time $h=20$, the first time occurs at $t = \dfrac{7}{6}$, the second at $\dfrac{11}{6}$, the third at $\dfrac{7}{6}+2(1)$ and the fourth at $\dfrac{11}{6}+2(1) = \dfrac{23}{6}$ seconds.
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  3. #3
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    Re: ball height

    thank you very much. i understand it.
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