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    hard projectile motion problem

    9. A boy throws a ball with speed V m/s at an angle of 45 degrees to the horizontal.

    a) Derive expressions for the horizontal and vertical components of the displacement of the ball from the point of projection

    b) Hence show that the Cartesian equation of the path o the ball is y = x- gx^2/ (V^2)

    c) the boy is now standing on a hill inclined at an angle A to the horizontal. He throws the ball at the same angle of elevation of 45 degress and at the same speed of V m/s. If he can throw the ball 60 metres down the hill but only 30 metres up the hill, use the result in part (b) to show that

    tan A = 1 - 30gcosA/V^2 = 60gcosA/V^2 -1

    and hence that tan A = 1/3

    I know how to do a,b but don't know how to do C, please help me
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  2. #2
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    Quote Originally Posted by hkdrmark View Post
    9. A boy throws a ball with speed V m/s at an angle of 45 degrees to the horizontal.

    a) Derive expressions for the horizontal and vertical components of the displacement of the ball from the point of projection

    b) Hence show that the Cartesian equation of the path o the ball is y = x- gx^2/ (V^2)

    c) the boy is now standing on a hill inclined at an angle A to the horizontal. He throws the ball at the same angle of elevation of 45 degress and at the same speed of V m/s. If he can throw the ball 60 metres down the hill but only 30 metres up the hill, use the result in part (b) to show that

    tan A = 1 - 30gcosA/V^2 = 60gcosA/V^2 -1

    and hence that tan A = 1/3

    I know how to do a,b but don't know how to do C, please help me
    A is assumed to be smaller than 45 degrees.

    Up the slope: Position measured from from point of projection.

    When the ball lands (on the slope): x = 30 \cos A and y = 30 \sin A.

    Substitute into to (b):

    30 \sin A = 30 \cos A - \frac{900 g \cos^2 A}{V^2}

     \Rightarrow \sin A = \cos A - \frac{30 g \cos^2 A}{V^2}

    \Rightarrow \tan A = 1 - \frac{30 g \cos A}{V^2} .... (1)


    Down the slope: Position measured from point of projection.

    When the ball lands (on the slope, you're meant to assume): x = 60 \cos A and y = - 60 \sin A.

    Substitute into (b):

    -60 \sin A = 60 \cos A - \frac{3600 g \cos^2 A}{V^2}

     \Rightarrow \sin A = - \cos A + \frac{60 g \cos^2 A}{V^2}

    \Rightarrow \tan A = -1 + \frac{60 g \cos A}{V^2} .... (2)


    Equate (1) and (2) and solve for \cos A.

    Substitute the expression for \cos A into (1) and simplify to get the required answer for \tan A.
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