Question

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second.

y=-16x^2+64x+89

Answer 1

Explanation:

I assume that "ground" is at 0 ft height. which is in an actual scenario not airways the case.

y = -16x² + 64x + 89

shows us that the tower is 89 ft tall (the result for x = 0, at the start).

anyway, if the original assumption is correct, then we need to solve

0 = -16x² + 64x + 89

the general solution for such a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/)2a)

in our case

a = -16

b = 64

c = 89

x = (-64 ± sqrt(64² - 4×-16×89))/(2×-16) =

= (-64 ± sqrt(4096 + 5696))/-32 =

= (-64 ± sqrt(9792))/-32

x1 = (-64 + 98.95453501...)/-32 = -1.092329219... s

x2 = (-64 - 98.95453501...)/-32 = 5.092329219... s

the negative solution for time is but useful here (it would be the time calculated back to ground at the start).

so, x2 is our solution.

the rocket hits the ground after about 5.09 seconds.