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A ball is thrown from a point 1.38 m above the ground. The initial velocity is 19.9 m/s at an angle of 33.8° above the horizontal. Find the maximum height of the ball above the ground.
Calculate the speed of the ball at the highest point in the trajectory.

  • physics -

    looking at energy, at the highest point, the ball will have no vertical KE, so total energy will be initial PE+initialKE

    topenergy=initial energy
    1/2 mVhoriz^2+mg(height)=1/2 m19.9^2+mg(1.38)

    where Vhoriz=19.9cosine33.8deg
    solve for height.

    and Vhoriz is the speed at the hightest point.

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