An astronaut on the moon drops a feather from rest (v = 0). The acceleration due to gravity is 1.67 m/s​ 2​ . If the feather begins 2 meters

Question

An astronaut on the moon drops a feather from rest (v = 0). The acceleration due to gravity is 1.67 m/s​ 2​ . If the feather begins 2 meters above the moon’s surface, what will be its final position after falling for 1.5 seconds?

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Khoii Minh 2 months 2021-07-21T10:11:53+00:00 1 Answers 2 views 0

Answers ( )

    0
    2021-07-21T10:13:22+00:00

    Given :

    Initial velocity , u = 0 m/s .

    Acceleration due to gravity on moon , g_m=1.67\ m/s^2 .

    Height , h = 2 m .

    To Find :

    Final position after falling for 1.5 seconds .

    Solution :

    We know , by equation of motion :

    s=ut+\dfrac{at^2}{2}

    Here , a = g_m .

    So , equation will transform by :

    s=ut+\dfrac{g_mt^2}{2}\\\\s=0+\dfrac{1.67\times 1.5^2}{2}\ m\\\\s=1.88\ m

    Therefore , the height form moon’s surface is 1.88 m .

    Hence , this is the required solution .

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