Chapter 8.4, Problem 42E

(a)

To determine

To find: the time the bullet hits the ground

Answer to Problem 42E

The bullet hits the ground after $64\sec onds$ .

Explanation of Solution

Given:

Initial speed of $2048ft/s$

Angle = $30°$ to the horizontal.

Calculation:

A bullet is fired with angle of elevation $\alpha =30°$ and its initial speed is $v_0=2048ft/s$ .

Find the time $t$ at which the bullet hits the ground.

The position of the bullet after $t$ seconds is given by the following parametric equations.

  $\begin{array}{l}x=\left(v_0\cos \alpha \right)t\\ y=\left(v_0\sin \alpha \right)t-16t^2\end{array}$

Here, $x,y$ are measured in feet.

The path of the bullet is shown below.

  

Substitute $v_0=2048ft/s,\alpha =30°$ in the above parametric equations.

Then, the position of the bullet after $t$ seconds is given by the following parametric equations.

  $\begin{array}{l}x=\left(2048\cos 30°\right)t\\ y=\left(2048\sin 30°\right)t-16t^2\end{array}$

Impact occurs when the y-component of the parametric equations is zero.

  $y=0$

  $\begin{array}{l}y=0\\ \left(2048\sin 30°\right)t-16t^2=0\\ \left(2048\left(\frac{1}{2}\right)\right)t-16t^2=0\\ 1024t-16t^2=0\\ 16t\left(64-t\right)=0\end{array}$

Either $t=0$ or $t=64$

Here $t=0$ gives the initial position of the bullet.

So, take $t=64$ .

Therefore, the bullet hits the ground after $64\sec onds$ .

Conclusion:

Therefore, the bullet hits the ground after $64\sec onds$ .

(b)

To determine

To find: the distance taken by the bullet to hit the ground

Answer to Problem 42E

The bullet hit the ground about $113512ft$ away.

Explanation of Solution

Calculation:

Find the horizontal distance $d$ that the bullet travelled.

The horizontal distance $d$ is the value of $x$ when $y=0$ .

From part $(a)$ , $y-component$ of the parametric equations is zero only when $t=64$ seconds.

So, find the value of $x$ by substituting $t=64$ in $x-component$ of the parametric equations of the bullet.

  $\begin{array}{l}x=\left(2048\cos 30°\right)t\\ x=\left(2048\cos 30°\right)\left(64\right)\left[Substitute\text{ t=64}\right]\\ =\left(2048\left(\frac{\sqrt{3}}{2}\right)\right)\left(64\right)\\ =65536\sqrt{3}\\ \approx 113511.6818\\ \approx 113512\end{array}$

Therefore, the bullet hit the ground about $113512ft$ away.

Conclusion:

Therefore, the bullet hit the ground about $113512ft$ away.

(c)

To determine

To find: the maximum height attained by the bullet

Answer to Problem 42E

The maximum height attained by the bullet is $16384ft$ .

Explanation of Solution

Calculation:

Find the maximum height attained by the bullet.

The position of the bullet after $t$ seconds is given by the following parametric equations.

  $\begin{array}{l}x=\left(2048\left(\frac{\sqrt{3}}{2}\right)\right)t\\ y=\left(2048\left(\frac{1}{2}\right)\right)t-16t^2\\ or\\ x=\left(1024\sqrt{3}\right)t\mathrm{.......}(1)\\ y=1024t-16t^2\mathrm{.........}(2)\end{array}$

First show that the path of the projectile is a parabola by eliminating the parameter $t$ .

From the equation $(1)$ , $t=\frac{x}{1024\sqrt{3}}$

Substitute $t=\frac{x}{1024\sqrt{3}}$ in the equation $(2)$ .

  $\begin{array}{l}y=1024t-16t^2\\ y=1024\left(\frac{x}{1024\sqrt{3}}\right)-16{\left(\frac{x}{1024\sqrt{3}}\right)}^2\\ y=\frac{x}{\sqrt{3}}-\frac{x^2}{196608}\end{array}$

So, the equation $y=\frac{x}{\sqrt{3}}-\frac{x^2}{196608}$ represents a parabola.

Compare the equation $y=\frac{x}{\sqrt{3}}-\frac{x^2}{196608}$ with the standard form of the parabola.

  $y=a{x}^2+bx+c$

Here, $a=-\frac{1}{196608},b=\frac{1}{\sqrt{3}}$ and $c=0$ .

The highest point on a parabola has $x-coordinate$

  $x=-\frac{b}{2a}$ .

So, $x-coordinates$ of the highest point on the parabola is,

  $\begin{array}{l}x=-\frac{b}{2a}\\ =-\frac{\left(\frac{1}{\sqrt{3}}\right)}{2\left(\frac{-1}{196608}\right)}\\ =\frac{98304}{\sqrt{3}}\end{array}$

Find the $y-coordinate$ at $x=\frac{98304}{\sqrt{3}}$

  $\begin{array}{l}y=\frac{x}{\sqrt{3}}-\frac{x^2}{196608}\\ =\frac{\left(\frac{98304}{\sqrt{3}}\right)}{\sqrt{3}}-\frac{{\left(\frac{98304}{\sqrt{3}}\right)}^2}{196608}\\ =\frac{98304}{3}-\frac{{98304}^2}{\left(196608\right)3}\\ =16384\end{array}$

Therefore, the maximum height attained by the bullet is $16384ft$ .

Conclusion:

Therefore, the maximum height attained by the bullet is $16384ft$ .