An object is launched from a platform. Its height (in meters), x 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? seconds

Find when h(x) equals 0: First, we need to find when h(x) equals 0, because that's when the object will hit the ground. So we set the equation to 0 and solve for x: 0 = -5(x-4)^(2) + 180Isolate the squared term: Now, let's isolate the squared term: -5(x-4)^(2) = -180Divide by -5: Divide both sides by -5 to simplify: (x-4)^(2) = 36Take square root: Take the square root of both sides to solve for x-4: x-4 = ±6Solve for x: Now, we solve for x by adding 4 to both possible values of x-4: x = 4 + 6 or x = 4 - 6 x = 10 or x = -2Discard negative solution: Since time can't be negative, we discard x = -2 and keep x = 10. So, the object will hit the ground after 10 seconds.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

An object is launched from a platform.
Its height (in meters),
x 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?
seconds

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

View step-by-step help

Home

Math Problems

Algebra 1

Solve quadratic equations: word problems

Full solution

Q. An object is launched from a platform.

\newline

Its height (in meters),

\newline

x

seconds after the launch, is modeled by:

\newline

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

\newline

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

\newline

\text{seconds}

Find when $h(x)$ equals $0$ : First, we need to find when $h(x)$ equals $0$ , because that's when the object will hit the ground. $\newline$ So we set the equation to $0$ and solve for $x$ : $\newline$ $0 = -5(x-4)^{2} + 180$
Isolate the squared term: Now, let's isolate the squared term: $\newline$ $-5(x-4)^{2} = -180$
Divide by $-5$ : Divide both sides by $-5$ to simplify: $\newline$ $(x-4)^{2} = 36$
Take square root: Take the square root of both sides to solve for $x-4$ : $x-4 = \pm 6$
Solve for x: Now, we solve for $x$ by adding $4$ to both possible values of $x-4$ : $x = 4 + 6$ or $x = 4 - 6$ $x = 10$ or $x = -2$
Discard negative solution: Since time can't be negative, we discard $x = -2$ and keep $x = 10$ . So, the object will hit the ground after $10$ seconds.