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    If I have an smooth inclined plane of angle \alpha to the horizontal, and a smooth particle sitting at the bottom of the slope. I give the particle an impulse up the slope at angle \theta to the horizontal. Show that the landing angle is \tan \phi=\frac {1}{\cot \theta-2\tan \alpha}.

    I know that \tan \phi=-v_y/v_x, and I have the values them in the textbook as standard results. I can't seem to simplify that though...
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    (Original post by bbrain)
    If I have an smooth inclined plane of angle \alpha to the horizontal, and a smooth particle sitting at the bottom of the slope. I give the particle an impulse up the slope at angle \theta to the horizontal. Show that the landing angle is \tan \phi=\frac {1}{\cot \theta-2\tan \alpha}.

    I know that \tan \phi=-v_y/v_x, and I have the values them in the textbook as standard results. I can't seem to simplify that though...
    Calculate the time at landing
    v_0 \sin \theta \cdot t-\frac{g}{2}t^2=v_0 \cos \theta \cdot t \cdot \tan \alpha
    write down v_y and v_x at this moment
    The angle of the velocity vector to the horizontal here is
    \tan \beta =\frac{v_y}{v_x}
    THe landing angle to the slope
    \tan \phi =\tan(\beta -\alpha)=\frac{\tan \beta-\tan \alpha}{1+\tan \beta \cdot \tan \alpha}
 
 
 
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