Answer
a) The instantaneous velocity of the debris at $\frac{1}{2}$ seconds after the explosion is $56\text{ feet per second}$ and the instantaneous velocity $4$ seconds after the explosion is $-56\text{ feet per second}$.
b) The instantaneous velocity just before the debris hits the ground is $-72\text{ feet per second}$.
Work Step by Step
(a)
Consider that the function $s\left( t \right)=-16{{t}^{2}}+72t$ describes the height of the debris above the ground in feet, t seconds after the explosion with an initial velocity of $72\text{ feet per second}$.
Compute the derivative of $s\left( t \right)=-16{{t}^{2}}+72t$ 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 $s\left( t \right)=-16{{t}^{2}}+72t$.
So, $s\left( t+h \right)=-16{{\left( t+h \right)}^{2}}+72\left( t+h \right)$
Put value of $s\left( t+h \right)$ in $s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h}$ ,
$\begin{align}
& s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( -16{{\left( t+h \right)}^{2}}+72\left( t+h \right) \right)-\left( -16{{t}^{2}}+72t \right)}{h}
\end{align}$
Now, simplify ${{\left( t+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$
$\begin{align}
& s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( -16\left( {{t}^{2}}+{{h}^{2}}+2th \right)+72\left( t+h \right) \right)-\left( -16{{t}^{2}}+72t \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{-16{{t}^{2}}-16{{h}^{2}}-32th+72t+72h+16{{t}^{2}}-72t}{h}
\end{align}$
Combine the like terms in the numerator, then divide the numerator and denominator by h; this gives,
$s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( -32t-16h+72 \right)$
Apply the limits,
$s'\left( t \right)=\left( -32t+72 \right)$
Now, substitute $t=\frac{1}{2}$ in $s'\left( t \right)$ to compute the instantaneous velocity at $\frac{1}{2}$ second after the explosion.
$\begin{align}
& s'\left( \frac{1}{2} \right)=\left( -32\left( \frac{1}{2} \right)+72 \right) \\
& =-16+72 \\
& =56
\end{align}$
Thus, the instantaneous velocity $\frac{1}{2}$ second after the explosion is $56\text{ feet per second}$.
Now, substitute $t=4$ in $s'\left( t \right)$ to compute the instantaneous velocity at $4$ second after the explosion.
$\begin{align}
& s'\left( 4 \right)=\left( -32\left( 4 \right)+72 \right) \\
& =-128+72 \\
& =-56
\end{align}$
Thus, the instantaneous velocity $4$ seconds after the explosion is $-56\text{ feet per second}$.
(b)
Consider that the function $s\left( t \right)=-16{{t}^{2}}+72t$ 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), $s'\left( t \right)=\left( -32t+72 \right)$
The instantaneous velocity has to be calculated just before the debris hits the ground. So, set $s\left( t \right)=0$
$\begin{align}
& -16{{t}^{2}}+72t=0 \\
& 8t\left( 2t-9 \right)=0
\end{align}$
Now, put the factors $8t,\left( 2t-9 \right)$ equal to zero and solve for t:
$\begin{align}
& 8t=0 \\
& t=0
\end{align}$
Or,
$\begin{align}
& \left( 2t-9 \right)=0 \\
& 2t=9 \\
& t=\frac{9}{2}
\end{align}$
Now, substitute $t=\frac{9}{2}$ in $s'\left( t \right)$ to compute the instantaneous velocity at $\frac{9}{2}$ second after the explosion.
$\begin{align}
& s'\left( \frac{9}{2} \right)=-32\left( \frac{9}{2} \right)+72 \\
& =-144+72 \\
& =-72
\end{align}$
Thus, the instantaneous velocity just before the debris hits the ground is $-72\text{ feet per second}$.
