Answered

Paige launched a ball using a catapult she built. The height of the ball (in meters above the ground) [tex]\( t \)[/tex] seconds after launch is modeled by:
[tex]\[ h(t) = -5t^2 + 40t \][/tex]

Paige wants to know when the ball will hit the ground.

1. Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation.
[tex]\[ h(t) = -5t(t - 8) \][/tex]

2. How many seconds after launch does the ball hit the ground?
[tex]\[ \boxed{\phantom{}8\phantom{} \text{seconds}} \][/tex]



Answer :

GinnyAnswer
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.

Other Questions