Answer
a) The instantaneous velocity of the ball $1$ second after being hit is $32\text{ feet per second}$ and the instantaneous velocity of the ball $3$ seconds after being hit is $-32\text{ feet per second}$.
b) The ball reaches maximum height $2\text{ seconds}$ after being hit and the maximum height that the ball reaches is $68\text{ feet}$.
Work Step by Step
(a)
Consider that the function $s\left( t \right)=-16{{t}^{2}}+64t+4$ describes the height of the ball above the ground in feet, t seconds after being hit with an initial height of $4$ feet and an initial velocity of $64\text{ feet per second}$.
Compute the derivative of $s\left( t \right)=-16{{t}^{2}}+64t+4$ 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 $s\left( t \right)=-16{{t}^{2}}+64t+4$.
So, $s\left( t+h \right)=-16{{\left( t+h \right)}^{2}}+64\left( t+h \right)+4$
Put value of $s\left( t+h \right)$ in $s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h}$ ,
$\begin{align}
& s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( -16{{\left( t+h \right)}^{2}}+64\left( t+h \right)+4 \right)-\left( -16{{t}^{2}}+64t+4 \right)}{h}
\end{align}$
Now, simplify ${{\left( t+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$
$\begin{align}
& s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( -16\left( {{t}^{2}}+{{h}^{2}}+2th \right)+64\left( t+h \right)+4 \right)-\left( -16{{t}^{2}}+64t+4 \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{-16{{t}^{2}}-16{{h}^{2}}-32th+64t+64h+4+16{{t}^{2}}-64t-4}{h}
\end{align}$
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 being hit.
$\begin{align}
& s'\left( 1 \right)=\left( -32\left( 1 \right)+64 \right) \\
& =-32+64 \\
& =32
\end{align}$
Thus, the instantaneous velocity $1$ seconds after being hit is $32\text{ feet per second}$.
Now, substitute $t=3$ in $s'\left( t \right)$ to compute the instantaneous velocity at $3$ seconds after being hit.
$\begin{align}
& s'\left( 3 \right)=\left( -32\left( 3 \right)+64 \right) \\
& =-96+64 \\
& =-32
\end{align}$
Thus, the instantaneous velocity $3$ seconds after being hit is $-32\text{ feet per second}$.
(b)
Consider that the ball has zero instantaneous velocity and the function $s\left( t \right)=-16{{t}^{2}}+64t+4$ describes the height of the ball above the ground in feet, t seconds after being hit with an initial height of $4$ feet and an initial velocity of $64\text{ feet per second}$.
From part (a), $s'\left( t \right)=\left( -32t+64 \right)$
The maximum height has to be calculated. So, set $s'\left( t \right)$ equal to zero.
$\begin{align}
& \left( -32t+64 \right)=0 \\
& 32t=64 \\
& t=2
\end{align}$
Thus, the ball reaches maximum height $2$ seconds after being hit.
Now, substitute $t=2$ in $s\left( t \right)$ to compute the maximum height at $2$ seconds after being hit.
$\begin{align}
& s\left( 2 \right)=-16{{\left( 2 \right)}^{2}}+64\left( 2 \right)+4 \\
& =-64+128+4 \\
& =68
\end{align}$
Thus, the maximum height that the ball reaches is $68\text{ feet}$.
