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28 October, 12:04

The equation of projectile is y = ax - bx2. its horizontal range is?

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  1. 28 October, 12:36
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    Y = a x - b x^2

    Range is a/b

    y = tan Ф x - g x² / 2 u² cos² Ф

    tan Ф = a - equation 1

    b = g / 2u² cos² Ф so u² cos² Ф = g / 2b - equation 2

    R = u cos Ф * 2 * u sin Ф / g = 2/g sinФ u² cos Ф

    = 2 / g tan Ф u² cos² Ф by using equation 1 and equation 2

    = (2 / g) a (g / 2b) = a / b
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