Let's address the problem step-by-step:
### 1. Rewrite the Function in Factored Form
The height of the ball as a function of [tex]\( t \)[/tex] (the time in seconds) is given by:
[tex]\[ h(t) = -5t^2 + 40t \][/tex]
To rewrite this function in a factored form, we factor out the greatest common factor from the terms in the quadratic equation.
First, factor out the common factor, which is [tex]\(-5t\)[/tex]:
[tex]\[ h(t) = -5t^2 + 40t = -5t(t - 8) \][/tex]
So, the function [tex]\( h(t) \)[/tex] in factored form is:
[tex]\[ h(t) = -5t(t - 8) \][/tex]
### 2. Determine When the Ball Hits the Ground
To find when the ball hits the ground, we need to solve for [tex]\( t \)[/tex] when the height [tex]\( h(t) \)[/tex] is equal to 0:
[tex]\[ -5t(t - 8) = 0 \][/tex]
This equation is satisfied when either factor is zero:
1. [tex]\( -5t = 0 \)[/tex]
2. [tex]\( t - 8 = 0 \)[/tex]
Solving these:
1. [tex]\( t = 0 \)[/tex]
2. [tex]\( t = 8 \)[/tex]
The two solutions are [tex]\( t = 0 \)[/tex] and [tex]\( t = 8 \)[/tex] seconds.
The solution [tex]\( t = 0 \)[/tex] represents the time the ball is launched. Therefore, the time when the ball hits the ground after it has been launched is:
[tex]\[ t = 8 \text{ seconds} \][/tex]
### Summary
To summarize, we have rewritten the function in factored form:
[tex]\[ h(t) = -5t(t - 8) \][/tex]
The ball hits the ground [tex]\( \boxed{8} \)[/tex] seconds after launch.
