An object is launched from a platform. Its height (in meters), xx seconds after the launch, is modeled by: h (x) = -5 (x-4) ^2+180 How many seconds after being launched will the object hit the ground?

x=10

Step-by-step explanation:

h (x) = -5 (x-4) ^2+180

Set y = 0 to find when it hits the ground

0 = - 5 (x-4) ^2+180

Subtract 180 from each side

-180 = -5 (x-4) ^2+180-180

-180 = -5 (x-4) ^2

Divide by - 5

-180/-5 = -5/-5 (x-4) ^2

36 = (x-4) ^2

Take the square root of each side

±sqrt (36) = sqrt ((x-4) ^2)

±6 = (x-4)

Add 4 to each side

4±6 = (x-4) + 4

4±6=x

x = 10 or - 2

But time cannot be negative so x=10

Answer:100 meters

Step-by-step explanation:h (0) = -5 (0-4) ^2+180

=-5 (16) + 180

=100