Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 13

Suppose that a ball is thrown in the air with a velocity of $40 ft/s$. Its height (in feet) after $t$ seconds is given by $y = 40 t - 16 t^2$ .
Find the velocity when $t = 2$.
From the definition of instantaneous velocity,

$
\displaystyle
\nu(a)= \lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}\\
$



$
\begin{equation}
\begin{aligned}
\qquad
f(t)
& = 40t - 16t^2\\
\nu(a)& = \lim \limits_{h \to 0} \frac{40(a+h)-16(a+h)^2 - [40(a) - 16(a^2)]}{h} \\
\nu(a)& = \lim \limits_{h \to 0} \frac{\cancel{40a}+40h-\cancel{16a^2}+32ah+16h^2\cancel{-40a}+\cancel{16a^2}}{h}\\
\nu(a)& = \lim \limits_{h \to 0} \frac{40h-32ah-16h^2}{h}\\
\nu(a)& = \lim \limits_{h \to 0} \frac{\cancel{h}(40-32a-16h)}{\cancel{h}}\\
\nu(a)& = \lim \limits_{h \to 0} (40-32a-16h)\\
\nu(a)& = 40-32a-16(0)\\
\nu(a)& = 40 - 32a
\end{aligned}
\end{equation}
$


The velocity after $2s$ is $\displaystyle\nu(2) = 40 - 32 (2) = -24 \frac{ft}{s}$