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1 February, 12:02

A stone is thrown vertically upward with a speed of 35.0 m/s a) Howfast sit moving when it reaches a height of 13.0m How much time is required to reach this height? Wht s the maximum height it will reach? Explain why there are two answers for part 3

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  1. 1 February, 14:03
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    a)

    v = 31.15 m/s

    t = 0.393 sec

    h = 62.5 m

    Explanation:

    a)

    v₀ = initially speed of the stone = 35.0 m/s

    v = final speed of the stone = ?

    y = vertical displacement of the stone = 13.0 m

    a = acceleration due to gravity = - 9.8 m/s²

    using the equation

    v² = v₀² + 2 a y

    v² = 35² + 2 ( - 9.8) (13.0)

    v = 31.15 m/s

    t = time taken

    using the equation

    v = v₀ + a t

    31.15 = 35 + ( - 9.8) t

    t = 0.393 sec

    h = maximum height

    v' = final speed at the maximum height = 0 m/s

    using the equation

    v'² = v₀² + 2 a h

    0² = 35² + 2 ( - 9.8) h

    h = 62.5 m
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