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A ball is thrown from a point 1.2 m above the ground. The initial velocity is 19.8 m/s at an angle of 34.0° above the horizontal.

(a) Find the maximum height of the ball above the ground.

(b) Calculate the speed of the ball at the highest point in the trajectory.


    Break the velocty into vertical and horizontal components.

    Then, height=1.2 + vivertical*t-4.9t^2

    and hmax occurs at the t when
    Vv=0=vivertical -9.8t

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