Chapter 9.4, Problem 64E

a.

To determine

To write a set of parametric equations that model the path of the arrow.

Answer to Problem 64E

  $x=\left(225\cos 15°\right)t$ and $y=3+\left(225\sin 15°\right)t-16t^2$ .

Explanation of Solution

Given: The archer releases the arrow at height of 5 ft above the ground. The arrow leaves the bow at angle of $15°$ with horizontal and at an initial speed of 225 ft per second.

Calculation:

The parametric equations for the path of the arrow can be given as:

  $x=\left(v_0\cos \theta \right)t=\left(225\cos 15°\right)t$ ,

  $y=h+\left(v_0\sin \theta \right)t-16t^2=3+\left(225\sin 15°\right)t-16t^2$ .

That is,

  $x=\left(225\cos 15°\right)t$ and $y=3+\left(225\sin 15°\right)t-16t^2$ .

b.

To determine

To find the distance the arrow travels before it hits the ground.

Answer to Problem 64E

  $802\text{ ft}$ .

Explanation of Solution

Given: The parametric equations for the path of the arrow is given as $x=\left(225\cos 15°\right)t$ and $y=3+\left(225\sin 15°\right)t-16t^2$ .

Calculation:

When arrow hits the ground, $y=0$ .

So,

  $\begin{array}{c}y=0\\ 3+\left(225\sin 15°\right)t-16t^2=0\\ 3+\left(225\cdot 0.2588\right)t-16t^2=0\\ 3+\left(\text{58}\text{.23}\right)t-16t^2=0\\ -16t^2+\left(\text{58}\text{.23}\right)t+3=0\\ t=-0.0508,t=3.69019\end{array}$

That means, the arrow will hit the ground after $t=3.69019\text{ sec}$ .

Therefore, the distance travelled by the arrow before hitting the ground is:

  $x=\left(225\cos 15°\right)t=\left(225\cos 15°\right)\cdot 3.69019=802.001\approx 802\text{ ft}$ .

c.

To determine

To graph the path of the arrow by using a graphing utility and approximate the maximum height of the arrow.

Answer to Problem 64E

55.988 ft.

Explanation of Solution

Given: The parametric equations for the path of the arrow is given as $x=\left(225\cos 15°\right)t$ and $y=3+\left(225\sin 15°\right)t-16t^2$ .

Graph:

The graph the parametric equations using graphing utility as shown below.

  

From the graph it can be observed that the maximum height attained by the arrow is 55.988 ft above the ground.

d.

To determine

To find the total time the arrow is in air.

Answer to Problem 64E

  $3.69019\text{ sec}$

Explanation of Solution

Given: The parametric equations for the path of the arrow is given as $x=\left(225\cos 15°\right)t$ and $y=3+\left(225\sin 15°\right)t-16t^2$ .

Calculation:

When arrow hits the ground, $y=0$ .

So,

  $\begin{array}{c}y=0\\ 3+\left(225\sin 15°\right)t-16t^2=0\\ 3+\left(225\cdot 0.2588\right)t-16t^2=0\\ 3+\left(\text{58}\text{.23}\right)t-16t^2=0\\ -16t^2+\left(\text{58}\text{.23}\right)t+3=0\\ t=-0.0508,t=3.69019\end{array}$

That means, the arrow will hit the ground after $t=3.69019\text{ sec}$ .

Therefore, the arrow is in air for $t=3.69019\text{ sec}$ .