## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1176: 76

#### Answer

The instantaneous velocity of the ball as it passes the roof is $-80\text{ feet per second}$.

#### Work Step by Step

Consider that the function $s\left( t \right)=-16{{t}^{2}}+80t+96$ describes the height of the ball above the ground in feet, t seconds after being hit with an initial height of $96\ \text{feet}$ and an initial velocity of $\text{80 feet per second}$. Compute the derivative of $s\left( t \right)=-16{{t}^{2}}+80t+96$ using the formula ${s}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( x+h \right)-s\left( x \right)}{h}$ as follows: To compute $s\left( t+h \right)$, substitute $x=t+h$ in the function $s\left( t \right)=-16{{t}^{2}}+64t+4$. \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}}+80\left( t+h \right)+96 \right)-\left( -16{{t}^{2}}+80t+96 \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)+80\left( t+h \right)+96 \right)-\left( -16{{t}^{2}}+80t+96 \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{-16{{t}^{2}}-16{{h}^{2}}-32th+80t+80h+96+16{{t}^{2}}-80t-96}{h} \end{align} Combine the like terms in the numerator, then divide the numerator and denominator by h; this gives, ${s}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( -32t-16h+80 \right)$ Apply the limits, ${s}'\left( t \right)=-32t+80$ Put $s\left( t \right)=96$ in the equation $s\left( t \right)=-16{{t}^{2}}+80t+96$ to find the time at which the ball passes the roof. \begin{align} & -16{{t}^{2}}+80t+96=96 \\ & 16{{t}^{2}}-80t=0 \\ & {{t}^{2}}-5t=0 \\ & t\left( t-5 \right)=0 \end{align} Now put the factors $t,\left( t-5 \right)$ equal to zero and solve for t: $t=0$ And \begin{align} & \left( t-5 \right)=0 \\ & t=5 \end{align} Now, $t=0$ will be the time when the ball was thrown up. So, the ball passes the roof on its way down after $5$ seconds. Now, substitute $t=5$ in ${s}'\left( t \right)$ to compute the instantaneous velocity at $5$ seconds after being thrown. ${s}'\left( t \right)=\left( -32t+80 \right)$ 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( t \right)=-32\left( 5 \right)+80 \\ & =-160+80 \\ & =-80 \end{align} Thus, the instantaneous velocity of the ball as it passes the roof is $-80\text{ feet per second}$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.