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4 April, 19:31

1a) An object is thrown up with a velocity of 18m/s from a height of 520m.

What's the object's position after 3s?

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  1. 4 April, 21:18
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    An object is thrown up from a height of 520m

    h_o = 520m

    Initial velocity of thrown is 18m/s

    u = 18m/s

    What is the position of the object after 3seconds

    t = 3s

    Acceleration due to gravity

    g = 9.81 m/s²

    Let calculated the height the object will reached when thrown from the top.

    Using equation of motion

    h = ut + ½gt²

    Since the body is thrown upward, it is acting against gravity then, gravity will be negative

    Then,

    h = ut - ½gt²

    h = 18 * 3 - ½ * 9.81 * 3²

    h = 54 - 44.145

    h = 9.855 m

    So, the body is above the top of the building at a distance of 9.855m

    So, the total distance from the bottom is

    Position = h + h_o

    x = 9.855 + 520

    x = 529.855m

    x ≈ 530m,

    The position of objects after 3seconds is 520m
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