Path of a projectile projected from the ground is given as y=ax-bx^2. find angle of projection, maximum height attained and horizontal range?

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  • 5 years ago
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    Soooo... under the assumption that ground is at y=0:

    Find the derivative of the path function. This tells you the slope of the function at any x.

    y' = a - (2b)x

    At x=0:

    y' = a

    The angle that the function makes with the x-axis at that point (projection angle) is tan^-1(a)

    For maximum height, slope = 0.

    a - (2b)x = 0

    x = a/(2b)

    So maximum height = ax-bx^2 = a(a/(2b)) - b(a/(2b))^2 = a^2/(2b) - a^2/(4b)

    Horizontal range is the non-zero value for x when y=0.

    y=0 ===> ax-bx^2 = 0 ===> x(a - bx) = 0 ===> x = a/b = range

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