A ball is placed in a machine that throws the ball up in the air. The table represents some points on the graph of a function that models the ball's distance from the ground with respect to the time since the ball has been th
rown.
\[
\begin{array}{|c|c|}
\hline
\text{Time (seconds)} & \text{Distance from Ground (feet)} \\
\hline
0 & 10 \\
1 & 20 \\
2 & 25 \\
3 & 20 \\
4 & 10 \\
\hline
\end{array}
\]
To find a function that models the ball's distance from the ground with respect to time, we can use polynomial interpolation. One common way to do this is to use Lagrange interpolation.
Let's define the Lagrange basis polynomials:
\[
\begin{aligned}
L_0(t) & = \frac{(t-1)(t-2)(t-3)(t-4)}{(0-1)(0-2)(0-3)(0-4)} = \frac{(t-1)(t-2)(t-3)(t-4)}{24} \\
L_1(t) & = \frac{(t-0)(t-2)(t-3)(t-4)}{(1-0)(1-2)(1-3)(1-4)} = -\frac{(t)(t-2)(t-3)(t-4)}{6} \\
L_2(t) & = \frac{(t-0)(t-1)(t-3)(t-4)}{(2-0)(2-1)(2-3)(2-4)} = \frac{(t)(t-1)(t-3)(t-4)}{2} \\
L_3(t) & = \frac{(t-0)(t-1)(t-2)(t-4)}{(3-0)(3-1)(3-2)(3-4)} = -\frac{(t)(t-1)(t-2)(t-4)}{4} \\
L_4(t) & = \frac{(t-0)(t-1)(t-2)(t-3)}{(4-0)(4-1)(4-2)(4-3)} = \frac{(t)(t-1)(t-2)(t-3)}{24} \\
\end{aligned}
\]
Now, we can write the interpolating polynomial as:
\[
P(t) = 10L_0(t) + 20L_1(t) + 25L_2(t) + 20L_3(t) + 10L_4(t)
\]
Expanding and simplifying, we get:
\[
P(t) = \frac{5}{3}t^3 - 10t^2 + \frac{65}{3}t + 10
\]
Therefore, the function that models the ball's distance from the ground with respect to time since the ball has been thrown is:
\[
f(t) = \frac{5}{3}t^3 - 10t^2 + \frac{65}{3}t + 10
\]