Answer
a) The instantaneous velocity $1$ second after the explosion is $32\text{ feet per second}$ and the instantaneous velocity $3$ seconds after the explosion is $-32\text{ feet per second}$.
b) The instantaneous velocity just before the debris hits the ground is $-64\text{ feet per second}$.
Work Step by Step
(a)
Consider that the function $s\left( t \right)=-16{{t}^{2}}+64t$ describes the height of the debris above the ground in feet, t seconds after the explosion with an initial velocity of $64\text{ feet per second}$.
Compute the derivative of $s\left( t \right)=-16{{t}^{2}}+64t$ using the formula $s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h}$ as follows:
To compute $s\left( t+h \right)$, substitute $t=t+h$ in the function $s\left( t \right)=-16{{t}^{2}}+64t$.
$s\left( t+h \right)=-16{{\left( t+h \right)}^{2}}+64\left( t+h \right)$
Put value of $s\left( t+h \right)$ in $s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h}$ ,
$\begin{align}
& s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( -16{{\left( t+h \right)}^{2}}+64\left( t+h \right) \right)-\left( -16{{t}^{2}}+64t \right)}{h}
\end{align}$
Now, simplify ${{\left( t+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$
$\begin{align}
& s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( -16\left( {{t}^{2}}+{{h}^{2}}+2th \right)+64\left( t+h \right) \right)-\left( -16{{t}^{2}}+64t \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{-16{{t}^{2}}-16{{h}^{2}}-32th+64t+64h+16{{t}^{2}}-64t}{h}
\end{align}$
Combine the like terms in the numerator, then divide the numerator and denominator by h; this gives,
$s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( -32t-16h+64 \right)$
Apply the limits,
$s'\left( t \right)=-32t+64$
Now, substitute $t=1$ in $s'\left( t \right)$ to compute the instantaneous velocity at $1$ second after the explosion.
$\begin{align}
& s'\left( 1 \right)=-32\left( 1 \right)+64 \\
& =-32+64 \\
& =32
\end{align}$
Thus, the instantaneous velocity $1$ second after the explosion is $32\text{ feet per second}$.
Now, substitute $t=3$ in $s'\left( t \right)$ to compute the instantaneous velocity at $3$ second after the explosion.
$\begin{align}
& s'\left( 3 \right)=-32\left( 3 \right)+64 \\
& =-96+64 \\
& =-32
\end{align}$
Thus, the instantaneous velocity $3$ seconds after the explosion is $-32\text{ feet per second}$.
(b)
Consider that the function $s\left( t \right)=-16{{t}^{2}}+64t$ describes the height of the debris above the ground in feet, t seconds after the explosion with an initial velocity of $64\text{ feet per second}$
From part (a), $s'\left( t \right)=\left( -32t+64 \right)$
The instantaneous velocity has to be calculated just before the debris hits the ground. So, set $s\left( t \right)=0$
$\begin{align}
& -16{{t}^{2}}+64t=0 \\
& {{t}^{2}}-4t=0 \\
& t\left( t-4 \right)=0
\end{align}$
Now, put the factors $t,\left( t-4 \right)$ equal to zero and solve for t:
$t=0$
Or
$\begin{align}
& \left( t-4 \right)=0 \\
& t=4
\end{align}$
Now, substitute $t=4$ in $s'\left( t \right)$ to compute the instantaneous velocity at $4$ second after the explosion.
$\begin{align}
& s'\left( 4 \right)=-32\left( 4 \right)+64 \\
& =-128+64 \\
& =-64
\end{align}$
Thus, the instantaneous velocity just before the debris hits the ground is $-64\text{ feet per second}$.
