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9 September, 00:59

gThe acceleration of gravity at the surface of Moon is 1.6 m/s2. A 5.0 kg stone thrown upward on Moon reaches a height of 20 m. (a) Find its initial velocity. (b) What is the time of flight to reach the max height

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  1. 9 September, 03:52
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    (a) 8 m/s

    (b) 5 s

    Explanation:

    (a)

    Using newton's equation of motion,

    v² = u²+2gs ... Equation 1

    Where v = final velocity, u = initial velocity, g = acceleration due to gravity at the surface of moon, s = height reached.

    make u the subject of the equation,

    u² = v²-2gs

    u = √ (v²-2gs) ... Equation 2

    Note: As the stone is thrown up, v = 0 m/s, g is negative

    Given: v = 0 m/s, s = 20 m, g = - 1.6 m/s²

    Substitute into equation 2

    u = √ (0-2*20*[-1.6])

    u = √64

    u = 8 m/s.

    (b)

    Using,

    v = u + gt

    Where t = time of flight to reach the maximum height.

    Make t the subject of the equation,

    t = (v-u) / g ... Equation 3

    Given: v = 0 m/s, u = 8 m/s, g = - 1.6 m/s²

    Substitute into equation 3

    t = (0-8) / -1.6

    t = - 8/-1.6

    t = 5 seconds.
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