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 Ho

Question

3 years ago

8

w many seconds after being launched will the object hit the ground?

2 answers:

densk [106] · Answer 1 · 2022-02-05T03:32:17+03:00

6 0

Answer:100 meters

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

=-5(16)+180

=100

kozerog [31] · Answer 2 · 2022-02-05T01:02:59+03:00

Answer:

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