math/pre-calc

"The roadrunner is standing under the edge of a cliff 570 feet above the ground. Wile E Coyote is 10 feet above the ground on a ledge below the roadrunner. Wile E. has purchased a spring catapult from ACME. The catapult will propel Wile E. with an upward velocity of 100 feet per second. The equation that gives the height, h, of Wile E. Coyote above the ground t seconds after he releases the catapult is given: h(t)=-4.9t^2+100t+10" When does the coyote reach the maximum height? I really cannot figure this one out. Thank You!

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asked by CPr
  1. tmax lies on the axis of symmetry of the parabola ... -b / 2a ... -100 / (2 * -4.9)

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    posted by R_scott
  2. First of all, your equation starting with -4.9t^2 + .... would be valid for metres and not for feet, on top of that the +100t would be the correct term for units in feet.
    All other units are in feet, what a messed-up question.
    So check on the wording of the question ....

    I will assume that all measurements are in metres.

    The max height will be obtained at the vertex of
    h = -4.9t^2 + 100t + 10
    the t of the vertex = -b/(2a) = -100/-9.8 = 10.204..
    So the max height above ground is obtained after 10.2 seconds

    at that t, h = -4.9(10.204.)^2 + 100(10.204..) + 10 = 520.204 m

    Since the cliff is 570 m high, as usual the coyote will fail and probably crash
    and in typical cartoon fashion, not hurt himself.

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    posted by Reiny

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