physics

A stone is thrown from the top of a building upward at an angle of 30.0∘to the horizontal and with an initial speed of 20.0 m/s. If the height of the building is 45.0 m, what is the speed in m/sec of the stone just before it strikes the ground ?

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  1. u = 20 cos 30 until it hits the ground.

    Vi = initial vertical velocity = +20 sin 30 = +10

    initial vertical displacement = 45
    final vertical displacement = 0

    0 = 45 + Vi t - 4.9 t^2
    solve for t

    v = Vi - 9.8 t
    solve for v

    speed = sqrt (u^2 + v^2)

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    posted by Damon

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