Physics

A ball is thrown from a point 1.0 m above the ground. The initial velocity is 19.6 m/s at an angle of 30 degrees 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.

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  1. max height is

    h = v^2 sin^2(θ)/2g
    = 19.6^2 * sin^2(30°)/(2*9.8)
    = 4.9m

    the speed at max height is just the horizontal component, so

    19.6cos(30°) = 16.97m/s

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  2. Thank you!! <3

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