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28 January, 02:33

A rock of mass m is thrown straight up into the air with initial speed |v0 | and initial position y = 0 and it rises up to a maximum height of y = h. A second rock with mass 2m (twice the mass of the original) is thrown straight up with an initial speed of 2|v0 |. What maximum height does the second rock reach?

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  1. 28 January, 04:05
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    Case 1:

    mass = m

    initial velocity = vo

    final velocity = 0

    height = y

    Use third equation of motion

    v² = u² - 2as

    0 = vo² - 2 g y

    y = vo² / 2g ... (1)

    Case 2:

    mass = 2m

    initial velocity = 2vo

    final velocity = 0

    height = y '

    Use third equation of motion

    v² = u² - 2as

    0 = 4vo² - 2 g y'

    y ' = 4vo² / 2g

    y' = 4 y

    Thus, the second rock reaches the 4 times the distance traveled by the first rock.
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