Question

An object is launched from a platform.
Its height (in meters),
\[x\] seconds after the launch, is modeled by
\[h(x)=-5(x+1)(x-9)\]
How many seconds after launch will the object hit the ground?
seconds

Answer 1

To find out when the object hits the ground, we set the height function \( h(x) \) equal to zero. The function is given by:

\[
h(x) = -5(x+1)(x-9)
\]

Setting \( h(x) \) to zero yields:

\[
-5(x+1)(x-9) = 0
\]

Since \(-5\) does not affect the roots, we can simplify the equation by dividing both sides by \(-5\):

\[
(x+1)(x-9) = 0
\]

Now, we can solve for \(x\) by setting each factor equal to zero:

1. \( x + 1 = 0 \)
\[
x = -1
\]

2. \( x - 9 = 0 \)
\[
x = 9
\]

The two solutions are \( x = -1 \) seconds and \( x = 9 \) seconds. Since negative time doesn't make sense in this context, we discard \( x = -1 \).

Therefore, the object will hit the ground \( 9 \) seconds after launch.

Thus, the answer is:

\[
\boxed{9} \text{ seconds}
\]