Chapter 3.4, Problem 64E

a.

To determine

To calculate: The time for the arrow to hit a target that is 3 feet above the ground.

Answer to Problem 64E

The required for the arrow to hit a target that is 3 feet above the ground is 0.35 seconds.

Explanation of Solution

Given information:

The initial height of the arrow = 5 feet

The height of the arrow = 3 feet

Calculation:

Consider the required time be t , the initial height be $h_0$ and the initial vertical velocity be $v_0$ .

So, $h_0=5$ , $v_0=0$ and $h=3$

The model of hitting of the arrow can be written as:

  $h=-16t^2+v_{0}t+h_0$

Substitute 0 for $v_0$ , 5 for $h_0$ and 3 for h in the above equation and simplify for t .

  $\begin{array}{c}3=-16t^2+\left(0\right)t+5\\ 16t^2=5-3\\ 16t^2=2\end{array}$

Rewrite the above equation as:

  $\begin{array}{c}16t^2=2\\ \frac{16t^2}{16}=\frac{2}{16}\\ t^2=\frac{1}{8}\end{array}$

Take square root of the above expression and simplify for t .

  $\begin{array}{c}\sqrt{t^2}=\sqrt{\frac{1}{8}}\\ t\approx \pm 0.35\end{array}$

Since, the value time cannot be negative.

Hence, the required for the arrow to hit a target that is 3 feet above the ground is 0.35 seconds.

b.

To determine

To determine: The method that is used to solve the above equation.

Answer to Problem 64E

The required method is taking square root of the both side of the equation.

Explanation of Solution

Given information:

The initial height of the arrow = 5 feet

The height of the arrow = 3 feet

Calculation:

Consider the required time be t , the initial height be $h_0$ and the initial vertical velocity be $v_0$ .

So, $h_0=5$ , $v_0=0$ and $h=3$

The model of hitting of the arrow can be written as:

  $h=-16t^2+v_{0}t+h_0$

Substitute 0 for $v_0$ , 5 for $h_0$ and 3 for h in the above equation and simplify for t .

  $\begin{array}{c}3=-16t^2+\left(0\right)t+5\\ 16t^2=5-3\\ 16t^2=2\end{array}$

Rewrite the above equation as:

  $\begin{array}{c}16t^2=2\\ \frac{16t^2}{16}=\frac{2}{16}\\ t^2=\frac{1}{8}\end{array}$

Take square root of the above expression and simplify for t .

  $\begin{array}{c}\sqrt{t^2}=\sqrt{\frac{1}{8}}\\ t\approx \pm 0.35\end{array}$

Since, the value time cannot be negative.

So, the arrow to hit a target that is 3 feet above the ground is 0.35 seconds.

Here, the equation $16t^2=2$ is in the form of $a{x}^2+c=0$ .

To solve the above equation, first remove the coefficient of $x^2$ as:

  $\begin{array}{c}a{x}^2+c=0\\ \frac{\cancel{a}{x}^2}{\cancel{a}}=\frac{-c}{a}\\ x^2=-\frac{c}{a}\end{array}$

Now, take square root on each side of the above equation to solve the equation.

  $\begin{array}{c}\sqrt{x^2}=\pm \sqrt{-\frac{c}{a}}\\ x=\pm \sqrt{-\frac{c}{a}}\end{array}$

Hence, the required method is taking square root of the both side of the equation.