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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.

  • PHYSICS - ,

    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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