## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 894: 45

#### Answer

a) The solution set for the system of equations involving $a,{{v}_{0}},{{s}_{0}}$ is $\left( -32,56,0 \right)$. b) The height of the ball is $0\text{ feet}$ after $3.5\text{ seconds}$. c) The maximum height is $49\text{ feet}$ and we attain it after $1.75\text{ seconds}$.

#### Work Step by Step

(a) Consider the provided graph . The point $\left( 1,40 \right)$ indicates the height $\left( s \right)$ of the ball is $40\text{ feet}$ after $1\text{ second}$. Similarly, $\left( 2,48 \right)$ indicates the height $\left( s \right)$ of the ball is $48\text{ feet}$ after $2\text{ seconds}$ and $\left( 3,24 \right)$ indicates the height $\left( s \right)$ of the ball is $40\text{ feet}$ after $3\text{ seconds}$. Now, substitute these values in the provided position function, $s\left( t \right)=\frac{1}{2}a{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$ For the point $\left( 1,40 \right)$ , \begin{align} & 40=\frac{1}{2}a\times {{(1)}^{2}}+{{v}_{0}}\times 1+{{s}_{0}} \\ & =\frac{1}{2}a+{{v}_{0}}+{{s}_{0}} \end{align} For the point $\left( 2,48 \right)$ , \begin{align} & 48=\frac{1}{2}a\times {{(2)}^{2}}+{{v}_{0}}\times 2+{{s}_{0}} \\ & =\frac{4}{2}a+2{{v}_{0}}+{{s}_{0}} \\ & =2a+2{{v}_{0}}+{{s}_{0}} \end{align} For the point $\left( 3,24 \right)$ , \begin{align} & 24=\frac{1}{2}a\times {{(3)}^{2}}+{{v}_{0}}\times 3+{{s}_{0}} \\ & =\frac{9}{2}a+3{{v}_{0}}+{{s}_{0}} \\ & =\frac{9}{2}a+3{{v}_{0}}+{{s}_{0}} \end{align} Thus, the system of equations is: \begin{align} & \frac{1}{2}a+{{v}_{0}}+{{s}_{0}}=40 \\ & 2a+2{{v}_{0}}+{{s}_{0}}=48 \\ & \frac{9}{2}a+3{{v}_{0}}+{{s}_{0}}=24 \\ \end{align} Now write the augmented matrix, $\left[ \begin{matrix} \frac{1}{2} & 1 & 1 & 40 \\ 2 & 2 & 1 & 48 \\ \frac{9}{2} & 3 & 1 & 24 \\ \end{matrix} \right]$ Now, reduce the matrix to row echelon form by using row operation ${{R}_{1}}\to 2{{R}_{1}}$ , $\left[ \begin{matrix} 1 & 2 & 2 & 80 \\ 2 & 2 & 1 & 48 \\ \frac{9}{2} & 3 & 1 & 24 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$ , $\left[ \begin{matrix} 1 & 2 & 2 & 80 \\ 0 & -2 & -3 & -112 \\ \frac{9}{2} & 3 & 1 & 24 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}-\frac{9}{2}{{R}_{1}}$ , $\left[ \begin{matrix} 1 & 2 & 2 & 80 \\ 0 & -2 & -3 & -112 \\ 0 & -6 & -8 & -336 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{-1}{2}{{R}_{2}}$ , $\left[ \begin{matrix} 1 & 2 & 2 & 80 \\ 0 & 1 & \frac{3}{2} & 56 \\ 0 & -6 & -8 & -336 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+6{{R}_{2}}$ , $\left[ \begin{matrix} 1 & 2 & 2 & 80 \\ 0 & 1 & \frac{3}{2} & 56 \\ 0 & 0 & 1 & 0 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}-\frac{3}{2}{{R}_{3}}$ , $\left[ \begin{matrix} 1 & 2 & 2 & 80 \\ 0 & 1 & 0 & 56 \\ 0 & 0 & 1 & 0 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-2{{R}_{3}}$ , $\left[ \begin{matrix} 1 & 2 & 0 & 80 \\ 0 & 1 & 0 & 56 \\ 0 & 0 & 1 & 0 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-2{{R}_{2}}$ , $\left[ \begin{matrix} 1 & 0 & 0 & -32 \\ 0 & 1 & 0 & 56 \\ 0 & 0 & 1 & 0 \\ \end{matrix} \right]$ Thus, $a=-32,{{v}_{0}}=56,{{s}_{0}}=0$ Hence, the solution set for the system of equations involving $a,{{v}_{0}},{{s}_{0}}$ is $\left( -32,56,0 \right)$. (b) Consider the provided function, $s\left( t \right)=\frac{1}{2}a{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$ Substitute $t=3.5$ in above equation, That is, \begin{align} & s\left( 3.5 \right)=\frac{1}{2}a{{(3.5)}^{2}}+{{v}_{0}}(3.5)+{{s}_{0}} \\ & =\frac{1}{2}\times 12.25\times a+3.5{{v}_{0}}+{{s}_{0}} \\ & =6.125a+3.5{{v}_{0}}+{{s}_{0}} \end{align} Now, substitute the values $a=-32,{{v}_{0}}=56,{{s}_{0}}=0$ obtained in part (a), \begin{align} & s\left( 3.5 \right)=6.125a+3.5{{v}_{0}}+{{s}_{0}} \\ & =6.125\times -32+3.5\times 56+0 \\ & =-196+196 \\ & =0 \end{align} Therefore, $s\left( 3.5 \right)=0$ Hence, the height of the ball is $0\text{ feet}$ after $3.5\text{ seconds}$. (c) The provided graph represents a parabola opening downwards. Therefore, the maximum height will be the vertex of parabola. The vertex of parabola is given by, $\left( -\frac{{{v}_{0}}}{2\left( \frac{1}{2}a \right)},f\left( -\frac{{{v}_{0}}}{2\left( \frac{1}{2}a \right)} \right) \right)$ Here, $-\frac{-{{v}_{0}}}{a}$ represents the time and $f\left( -\frac{{{v}_{0}}}{a} \right)$ represents the maximum height of the ball. Therefore, \begin{align} & -\frac{{{v}_{0}}}{a}=-\frac{56}{-32} \\ & =1.75 \end{align} Substitute $t=1.75$ in $s\left( t \right)=\frac{1}{2}a{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$ Thus, \begin{align} & s(1.75)=\frac{1}{2}(-32){{(1.75)}^{2}}+56(1.75)+0 \\ & =-49+98 \\ & =49 \end{align} Therefore, the ball attains the max height after $1.75\text{ seconds}$ and the maximum height is $49\text{ feet}$. Hence, the maximum height is $49\text{ feet}$ and we attain it after $1.75\text{ seconds}$

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