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Thread: Trigonometric help in mechanics question?

  1. #1
    Super Member fardeen_gen's Avatar
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    Trigonometric help in mechanics question?

    While solving a projectile motion problem, i found the elevation angle as
    θ = arccos ((2/3) + (v^2/3gR)) (which is correct according the solution). The next step requires tan θ, which I calculated as equal to √(5 - (v^4/g^2.R^2) - (4v^2/gR))/ (2 + (v^2/gR)). The solution in the text says that tan θ = - √(1 - (v^2/gR))/(√(2 + (v^2/gR))). Can anyone help me on what I am missing?
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  2. #2
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    Hello, fardeen_gen!

    I agree with your answer . . .


    While solving a projectile motion problem, i found the elevation angle as:
    $\displaystyle \theta \:=\:\arccos\left(\tfrac{2}{3} + \tfrac{v^2}{3gR}\right)$, (which is correct according the solution).

    The next step requires $\displaystyle \tan\theta$, which I calculated as equal to:
    . . $\displaystyle \frac{\sqrt{5 - \frac{v^4}{g^2R^2} - \frac{4v^2}{gR}}} {2 + \frac{v^2}{gR}}$
    The text says that: .$\displaystyle \tan\theta \:=\: -\frac{\sqrt{1 - \frac{v^2}{gR}}} {\sqrt{2 + \frac{v^2}{gR}}}$

    We have: .$\displaystyle \theta \:=\:\arccos\left(\frac{2gR+v^2}{3gR}\right) \quad\Rightarrow\quad\cos\theta \:=\:\frac{2gR + v^2}{3gR} \:=\:\frac{adj}{hyp} $

    Hence, $\displaystyle \theta$ is in a right triangle with: .$\displaystyle adj = 2gR + v^2,\;hyp = 3gR$

    . . Using Pythagorus, we have: .$\displaystyle opp \,=\,\sqrt{5g^2R^2 - 4gRv^2 - v^4}$


    Therefore: .$\displaystyle \tan\theta \;=\;\frac{opp}{adj} \;=\;\frac{\sqrt{5g^2R^2 - 4gRv^2 - v^4}}{2gR + v^2} $

    . . This is equivalent to your answer.

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