Chapter 2.7, Problem 68E

a .

To determine

To find: time when projectile will be back at the ground level.

Answer to Problem 68E

  $t=8\text{ sec}$

Explanation of Solution

Given:

Initial height $s_0=0$

Initial speed $v_0=128\text{ feet per second}$

Concept Used:

Position Equation:

  $s\left(t\right)=-16t^2+v_{0}t+s_0$ ….. (1)

Where s represents the height of an object (in feet), $v_0$ represents the initial velocity of the object (in feet per second), $s_0$ represents the initial height of the object (in feet), and t represents the time (in seconds).

When projectile will be back at the ground level, its height will be $s=0$

Substitute $s=0$ , $s_0=0$ and $v_0=128$ in equation (1) and solve it for t .

  $\begin{array}{c}-16t^2+128t+0=0\\ -16t^2+128t=0\\ -16t\left(t-8\right)=0\end{array}$

This implies that,

  $t=0$

Or

  $\begin{array}{c}t-8=0\\ t=8\end{array}$

So, the projectile will be back at the ground level after $t=8\text{ sec}$ .

b.

To determine

To find: time when projectile’s height will less than 128 feet.

Answer to Problem 68E

  $t<\text{1}\text{.17 sec}$ And $t>6.83\text{ sec}$ .

Explanation of Solution

Given:

Initial height $s_0=0$

Initial speed $v_0=128\text{ feet per second}$

Concept Used:

Position Equation:

  $s\left(t\right)=-16t^2+v_{0}t+s_0$ ….. (1)

Where s represents the height of an object (in feet), $v_0$ represents the initial velocity of the object (in feet per second), $s_0$ represents the initial height of the object (in feet), and t represents the time (in seconds).

Concept Used:

Quadratic Formula:

The zeros of the polynomial $f\left(x\right)=a{x}^2+bx+c$ , $a\ne 0$ are given by:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

When projectile’s height will less than 128 feet, $s<128$

Substitute $s<128$ , $s_0=0$ and $v_0=128$ in equation (1) and solve it for t .

  $\begin{array}{c}-16t^2+128t+0<128\\ -16t^2+128t-128<0\end{array}$

First step to solve the given inequality is to find the key numbers of the inequality. For that, find zeros of the polynomial $-16t^2+128t-128$ .

  $\begin{array}{c}-16t^2+128t-128=0\\ -16\left(t^2-8t+8\right)=0\\ t^2-8t+8=0\end{array}$

By quadratic formula, zeros of this polynomial are:

  $\begin{array}{c}t=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(1\right)\left(8\right)}}{2\left(1\right)}\\ =\frac{8\pm \sqrt{32}}{2}\\ \approx 1.17,6.83\end{array}$

Thus, the key numbers are $x=1.17$ and $x=6.83$ .

So, the inequality’s test intervals are $\left(-\infty ,1.17\right)$ , $\left(1.17,6.83\right)$ and $\left(6.83,\infty \right)$

In each test interval, choose a representative x -value and evaluate the polynomial.

Test-Interval	x -value	Polynomial Value	Conclusion
$\left(-\infty ,1.17\right)$	$x=0$	$-16{\left(0\right)}^2+128\left(0\right)-128=-128$	Negative
$\left(1.17,6.83\right)$	$x=5$	$-16{\left(5\right)}^2+128\left(5\right)-128=112$	Positive
$\left(6.83,\infty \right)$	$x=10$	$-16{\left(10\right)}^2+128\left(10\right)-128=-448$	Negative

The inequality is satisfied for all x -values in $\left(-\infty ,1.17\right)$ and $\left(6.83,\infty \right)$ .

This implies that the solution set of the inequality $-16t^2+128t-128<0$ is the interval $\left(-\infty ,1.17\right)$ and $\left(6.83,\infty \right)$ .

Conclusion:

The projectile’s height will less than 128 feet between $t<\text{1}\text{.17 sec}$ and $t>6.83\text{ sec}$ .