"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!
tmax lies on the axis of symmetry of the parabola ... -b / 2a ... -100 / (2 * -4.9)
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.
To find the time when the coyote reaches the maximum height, you need to find the vertex of the quadratic function representing the height, h(t), as a function of time, t.
The vertex of a quadratic function in the form of h(t) = at^2 + bt + c can be found using the formula t = -b/(2a). In this case, a = -4.9 and b = 100.
To find the time when the coyote reaches the maximum height, substitute these values into the formula:
t = -b/(2a)
= -100/(2(-4.9))
= -100/(-9.8)
= 10.2041 seconds (approximately)
So, the coyote reaches the maximum height at approximately 10.2041 seconds after releasing the catapult.