Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

An astronaut on Jupiter drops a rock straight downward from height of 1.20m. If the acceleration of gravity on Jupiter is 24.8m/s^2 what...?

is the speed of the rock just before it lands?

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  • 1 month ago
    Favorite Answer

    v = √(2gh)

    v = √(2•24.8•1.2) = 7.71 m/s

    falling object starting from rest

    h is height in meters, t is time falling in seconds,

    g is acceleration of gravity, usually 9.8 m/s²

    v is velocity in m/s

    h = ½gt²

    t = √(2h/g)

    v = √(2gh)

    h = v²/2g

    v = gt

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  • MyRank
    Lv 6
    1 month ago

    Given,

    Height (h) = 1.20m

    Acceleration due to gravity (g) = 24.8m/sec²

    Initial speed (u) = 0m/sec [∵ freely falling]

    Final speed (v) = ?

    Using kinematic relation:-

    v² = u² + 2gh

    = v² = (0m/sec)² + 2 x 24.8m/sec² x 1.20m

    = v = √2 x 24.8 x 1.20 = 7.71m/sec

    ∴ Final speed (v) = 7.71m/sec.

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  • 1 month ago

    a = gravitational acceleration on Jupiter = 24.8 m/s²

    d = distance of the rock's fall = 1.20 m

    v = final velocity of the rock = to be determined

    v = √(2ad)

    v = √(2·24.8 m/s²·1.20 m)

    v = 7.714920609 m/s ≈ 7.71 m/s

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