Question

An object is launched from a platform.

Its height (in meters), xxx seconds after the launch, is modeled by:
h(x)=-5(x-4)^2+180h(x)=−5(x−4)
2
+180h, left parenthesis, x, right parenthesis, equals, minus, 5, left parenthesis, x, minus, 4, right parenthesis, squared, plus, 180
How many seconds after being launched will the object hit the ground?

Answer 1

10

Explanation:

Ground level is where h = 0, so solve the equation ...

h(x) = 0

-5(x -4)^2 +180 = 0 . . . . substitute for h(x)

(x -4)^2 = 36 . . . . . . . . . . divide by -5, add 36

x -4 = 6 . . . . . . . . . . . . . . positive square root*

x = 10 . . . . . . add 4

The object will hit the ground 10 seconds after launch.

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* The negative square root also gives an answer that satisfies the equation, but is not in the practical domain. That answer would be x = -2. The equation is only useful for time at and after the launch time: x ≥ 0.