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21 October, 12:56

You throw an 18.0 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 17.0 m/s. Part A Use the work-energy principle to find the rock's speed just as it left the ground. Express your answer in meters per second. vground

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  1. 21 October, 13:22
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    Vi = 24.14 m/s

    Explanation:

    If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)

    Loss in K. E of the Rock = Gain in P. E of the Rock

    (1/2) (m) (Vi² - Vf²) = mgh

    Vi² - Vf² = 2gh

    Vi² = Vf² + 2gh

    Vi = √ (Vf² + 2gh)

    where,

    Vi = Rock's Speed as it left the ground = ?

    Vf = Final Speed = 17 m/s

    g = 9.8 m/s²

    h = height of rock = 15 m

    Therefore,

    Vi = √[ (17 m/s) ² + 2 (9.8 m/s²) (15 m) ]

    Vi = √583 m²/s²

    Vi = 24.14 m/s
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