Chapter 1.7, Problem 42E

(a)

To determine

To find: The time at which the bullet will hit the ground, the distance which the bullet travels and the maximum height reached by the bullet.

Answer to Problem 42E

The time taken by the bullet to hit the ground is $51\sec$ , the distance travelled by the bullet is $22083.65\text{m}$ and the maximum height attended by the bullet is $3188.77\text{m}$ .

Explanation of Solution

Given:

The initial velocity of the projectile is $v_O=500\text{m}/\text{s}$ .

The angle $\alpha$ is $\alpha =30°$ that is $\frac{\pi }{6}$ radians/

The parametric equations are,

  $\begin{array}{l}x=\left(v_O\cos \alpha \right)t\\ y=\left(v_O\sin \alpha \right)t-\frac{1}{2}g{t}^2\end{array}$

The gravity acceleration is $g=9.8\text{m}/{\text{s}}^2$

Calculation:

Consider the given parametric equations,

  $\begin{array}{l}x=\left(v_O\cos \alpha \right)t\\ y=\left(v_O\sin \alpha \right)t-\frac{1}{2}g{t}^2\end{array}$

Then,

  $\begin{array}{l}x=\left(500\cos \frac{\pi }{6}\right)t\\ y=\left(500\sin \frac{\pi }{6}\right)t-\frac{1}{2}g{t}^2\end{array}$

That is,

  $\begin{array}{l}x=250\sqrt{3}t\\ y=250t-\frac{1}{2}g{t}^2\end{array}$

Consider the time at which the bullet is the ground is at time when $y=0$ .

  $\begin{array}{l}250t-\frac{1}{2}g{t}^2=0\\ t=\frac{250\left(2\right)}{9.8\text{m}/{\text{s}}^2}\\ t=51\text{sec}\end{array}$

The distance between the gun and the bullet ground hit is,

  $250\sqrt{3}t=x$

Then,

  $\begin{array}{l}250\sqrt{3}t=x\\ 250\sqrt{3}\left(51\right)=x\\ x=22083.65\text{m}\end{array}$

Consider the maximum height of the bullet is,

  $y=250t-\frac{1}{2}gt$

Differentiate the above equation as,

  $\begin{array}{l}\frac{dy}{dt}=250-gt\\ 0=250-gt\\ t=\frac{250}{g}\sec \end{array}$

The maximum height of the bullet is at time $\frac{250}{g}\sec$ .

  $\begin{array}{c}y=250\left(\frac{250}{g}\sec \right)-\frac{1}{2}g{\left(\frac{250}{g}\sec \right)}^2\\ =\frac{62500}{2g}\\ =\frac{62500}{2\left(9.8\right)}\\ =3188.77\text{m}\end{array}$

(b)

To determine

To find: The graph for the path of the projectile for several other values of the angle $\alpha$ to see where the bullet hits the ground and summarize the findings.

Answer to Problem 42E

The graph for the path of the path of the projectile is shown in Figure 2 and it is observed that at $\alpha =45°$ the maximum height reached by the bullet is maximum

Explanation of Solution

Given:

The initial velocity of the projectile is $v_O=500\text{m}/\text{s}$ .

The angle $\alpha$ is $\alpha =30°$ that is $\frac{\pi }{6}$ radians/

The parametric equations are,

  $\begin{array}{l}x=\left(v_O\cos \alpha \right)t\\ y=\left(v_O\sin \alpha \right)t-\frac{1}{2}g{t}^2\end{array}$

The gravity acceleration is $g=9.8\text{m}/{\text{s}}^2$

Calculation:

Consider the parametric equations are,

  $\begin{array}{l}x=\left(v_O\cos \alpha \right)t\\ y=\left(v_O\sin \alpha \right)t-\frac{1}{2}g{t}^2\end{array}$

The graph for the above equations is shown in Figure 1

Figure 1

Consider the sketch for the curve when $\alpha =30°,45°,60°$ shown in Figure 2

Figure 2

It is observed that at $\alpha =45°$ the height of the bullet is maximum.

(c)

To determine

To find: The path of the bullet is parabolic.

Answer to Problem 42E

The equation for the path of the bullet is the equation of the parabola $y=x\tan \alpha -\frac{1}{2}g\frac{{\sec }^2\alpha }{{\left(v_O\right)}^2}{x}^2$ .

Explanation of Solution

Given:

The parametric equations are,

  $\begin{array}{l}x=\left(v_O\cos \alpha \right)t\\ y=\left(v_O\sin \alpha \right)t-\frac{1}{2}g{t}^2\end{array}$

Calculation:

Consider the parametric equations are,

  $\begin{array}{l}x=\left(v_O\cos \alpha \right)t\\ y=\left(v_O\sin \alpha \right)t-\frac{1}{2}g{t}^2\end{array}$ .... (I)

Then,

  $\begin{array}{c}t=\frac{x}{\left(v_O\cos \alpha \right)}\\ =\frac{x}{v_O}\sec \alpha \end{array}$

Substitute $\frac{x}{v_O}\sec \alpha$ for $t$ in the equation (I).

  $\begin{array}{c}y=\left(v_O\sin \alpha \right)\left(\frac{x}{v_O}\sec \alpha \right)-\frac{1}{2}g{\left(\frac{x}{v_O}\sec \alpha \right)}^2\\ =x\tan \alpha -\frac{1}{2}g\frac{{\sec }^2\alpha }{{\left(v_O\right)}^2}{x}^2\end{array}$

Thus, the equation for the path of the bullet is the equation of the parabola $y=x\tan \alpha -\frac{1}{2}g\frac{{\sec }^2\alpha }{{\left(v_O\right)}^2}{x}^2$ .