Answer
The value of $\frac{s\left( a+h \right)-s\left( a \right)}{h}$ is $-32a-16h+48$.
Work Step by Step
Consider the function $ s\left( t \right)=-16{{t}^{2}}+48t+160$,
Find the value of $\frac{s\left( a+h \right)-s\left( a \right)}{h}$ using the definition of the function.
$\begin{align}
& \frac{s\left( a+h \right)-s\left( a \right)}{h}=\frac{-16{{\left( a+h \right)}^{2}}+48\left( a+h \right)+160-\left( -16{{a}^{2}}+48a+160 \right)}{h} \\
& =\frac{-16\left( {{a}^{2}}+{{h}^{2}}+2ah \right)+48\left( a+h \right)+160+16{{a}^{2}}-48a-160}{h} \\
& =\frac{-16{{a}^{2}}-16{{h}^{2}}-32ah+48a+48h+160+16{{a}^{2}}-48a-160}{h} \\
& =\frac{-16{{h}^{2}}-32ah+48h}{h}
\end{align}$
Factoring out h from the numerator,
$\begin{align}
& \frac{s\left( a+h \right)-s\left( a \right)}{h}=\frac{h\left( -16h-32a+48 \right)}{h} \\
& =-32a-16h+48
\end{align}$
Thus, the value of $\frac{s\left( a+h \right)-s\left( a \right)}{h}$ is $-32a-16h+48$.
