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1 March, 10:47

An object is dropped near the surface of a planet such that the gravitational field at the object's location is 8 Nkg. If the object is thrown upward at a speed of 20 ms, what is the position of the object in relation to the position in which the object was released and thrown upward after 3s?

A - 96 m below the release position

B - 36 m below the release position

C-15 m above the release position

D - 24 m above the release position

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  1. 1 March, 11:02
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    D - 24 m above the release position

    Explanation:

    Given:

    v₀ = 20 m/s

    a = - 8 m/s²

    t = 3 s

    Find: Δx

    Δx = v₀ t + ½ at²

    Δx = (20 m/s) (3 s) + ½ (-8 m/s²) (3 s) ²

    Δx = 24 m
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