Sure, let's break down the function [tex]\( h(t) = -49t^2 + 98t \)[/tex] step by step to find the equivalent form that most easily reveals the length of time the ball is in the air.
1. Original Function: [tex]\( h(t) = -49t^2 + 98t \)[/tex]
2. Factoring the Function:
We want to factor this quadratic expression to find an equivalent form.
Notice that both terms, [tex]\(-49t^2\)[/tex] and [tex]\(98t\)[/tex], have a common factor. We can factor out the greatest common factor, which in this case is [tex]\( -49t \)[/tex].
[tex]\( h(t) = -49t^2 + 98t \)[/tex]
Factor out [tex]\(-49t\)[/tex]:
[tex]\[
h(t) = -49t(t - 2)
\][/tex]
3. Interpreting the Factored Form:
The factored form of the function is [tex]\( h(t) = -49t(t - 2) \)[/tex].
This form reveals the roots, or the solutions to the equation when [tex]\( h(t) = 0 \)[/tex]:
[tex]\[
-49t(t - 2) = 0
\][/tex]
The solutions to the equation are:
[tex]\[
t = 0 \quad \text{or} \quad t - 2 = 0 \implies t = 2
\][/tex]
These solutions represent the times when the ball is at height [tex]\( h(t) = 0 \)[/tex], which corresponds to the moments the ball is launched (at [tex]\( t = 0 \)[/tex]) and when it touches the ground again (at [tex]\( t = 2 \)[/tex]).
4. Conclusion:
The equivalent function that most easily reveals the length of time the ball is in the air is:
[tex]\[
h(t) = -49t(t - 2)
\][/tex]
This form shows that the ball is in the air for [tex]\( 2 \)[/tex] seconds, from the time it is launched until it returns to the ground.
So, the answer is:
[tex]\[
h(t) = -49t(t - 2)
\][/tex]
