Question

Curtain of death.A large metallic asteroid strikes Earth and quickly digs a crater into the rocky material below ground level by launching rocks upward and outward. The following table gives five pairs of launch speeds and angles (from the horizontal) for such rocks, based on a model of crater formation. (Other rocks, with intermediate speeds and angles, are also launched.) Suppose that you are at x=20 kmwhen the asteroid strikes the ground at time t=0nd position x=0(Fig. 4-52). (a) At t=20 s, what are the xand y coordinates of the rocks headed in your direction from launches A through E? (b) Plot these coordinates and then sketch a curve through the points to include rocks with intermediate launch speeds and angles.The curve should indicate what you would see as you look up into the approaching rocks.

Short answer

a) The x-and y-co-ordinates for different speed and angles for launches A to Eare:

A: (x, y) = (10.1, 0.556)km

B: (x, y) = (12.1, 1.51)km

C: (x, y) = (14.3, 2.68)km

D: (x, y) = (16.4, 3.99)km

E: (x, y) = (18.5, 5.53)km

b) The coordinates of the launches are plotted.

Step-by-step solution

The given data

$$\begin{gathered} \mathrm{A}->\mathrm{speed}=520\mathrm{m}/\mathrm{s},\mathrm{\theta }=14.0° \\ \mathrm{B}->\mathrm{speed}=630\mathrm{m}/\mathrm{s},\mathrm{\theta }=16.0° \\ \mathrm{C}->\mathrm{speed}=750\mathrm{m}/\mathrm{s},\mathrm{\theta }=18.0° \\ \mathrm{D}->\mathrm{speed}=870\mathrm{m}/\mathrm{s},\mathrm{\theta }=20.0° \\ \mathrm{E}->\mathrm{speed}=1000\mathrm{m}/\mathrm{s},\mathrm{\theta }=22.0° \end{gathered}$$

Understanding the concept of the projectile motion

Projectile motion is the motion of a particle that is launched with an initial velocity. During its flight, the particle’s horizontal acceleration is zero and its vertical acceleration is the free-fall acceleration (Upward is taken to be a positive direction.)

We can use the projectile motion concept to calculate the x- and y-coordinates for different speeds and angles.

Formulae:

The horizontal distance of the body in projectile motion, $\mathrm{x}={\mathrm{V}}_0\mathrm{cos\theta t}$ …(i)

The vertical distance of the body in projectile motion, $\mathrm{y}={\mathrm{V}}_0\mathrm{sin\theta }\mathrm{t}\frac{1}{2}{\mathrm{gt}}^2$ …(ii)

(a) Calculation of the coordinates of the launches from A to E at t = 20s

For launch of A,

The x-coordinate is given using equation (i) as:

$$\begin{gathered} \mathrm{x}=\left(520\mathrm{m}/\mathrm{s}\right)\cos \left(14°\right)\left(20\mathrm{s}\right) \\ =10.1\mathrm{km} \end{gathered}$$

The y-coordinate is given using equation (ii) as:

role="math" localid="1657020170294" $$\begin{gathered} \mathrm{y}=\left(520\mathrm{m}/\mathrm{s}\right)\sin \left(14°\right)\left(20\mathrm{s}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(20\mathrm{s}\right)}^2 \\ =0.556\mathrm{km} \end{gathered}$$

For launch of B,

The x-coordinate is given using equation (i) as:

$$\begin{gathered} \mathrm{x}=\left(630\mathrm{m}/\mathrm{s}\right)\cos \left(16°\right)\left(20\mathrm{s}\right) \\ =12.1\mathrm{km} \end{gathered}$$

The y-coordinate is given using equation (ii) as:

$$\begin{gathered} \mathrm{y}=\left(630\mathrm{m}/\mathrm{s}\right)\sin \left(16°\right)\left(20\mathrm{s}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(20\mathrm{s}\right)}^2 \\ =1.51\mathrm{km} \end{gathered}$$

For launch of C,

The x-coordinate is given using equation (i) as:

$$\begin{gathered} \mathrm{x}=\left(750\mathrm{m}/\mathrm{s}\right)\cos \left(18°\right)\left(20\mathrm{s}\right) \\ =14.3\mathrm{km} \end{gathered}$$

The y-coordinate is given using equation (ii) as:

$$\begin{gathered} \mathrm{y}=\left(750\mathrm{m}/\mathrm{s}\right)\sin \left(18°\right)\left(20\mathrm{s}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(20\mathrm{s}\right)}^2 \\ =2.68\mathrm{km} \end{gathered}$$

For launch of D,

The x-coordinate is given using equation (i) as:

$$\begin{gathered} \mathrm{x}=\left(870\mathrm{m}/\mathrm{s}\right)\cos \left(20°\right)\left(20\mathrm{s}\right) \\ =16.4\mathrm{km} \end{gathered}$$

The y-coordinate is given using equation (ii) as:

$$\begin{gathered} \mathrm{y}=\left(870\mathrm{m}/\mathrm{s}\right)\sin \left(20°\right)\left(20\mathrm{s}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(20\mathrm{s}\right)}^2 \\ =3.99\mathrm{km} \end{gathered}$$

For launch of E,

The x-coordinate is given using equation (i) as:

$$\begin{gathered} \mathrm{x}=\left(1000\mathrm{m}/\mathrm{s}\right)\cos \left(22°\right)\left(20\mathrm{s}\right) \\ =18.5\mathrm{km} \end{gathered}$$

The y-coordinate is given using equation (ii) as:

$$\begin{gathered} \mathrm{y}=\left(1000\mathrm{m}/\mathrm{s}\right)\sin \left(22°\right)\left(20\mathrm{s}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(20\mathrm{s}\right)}^2 \\ =5.53\mathrm{km} \end{gathered}$$

Hence, we get the x and y co-ordinates for all cases as:

A: (x, y) = (10.1, 0.556)km

B: (x, y) = (12.1, 1.51)km

C: (x, y) = (14.3, 2.68)km

D: (x, y) = (16.4, 3.99)km

E: (x, y) = (18.5, 5.53)km

(b) Calculation of plotting the graph using the coordinates

The plot of x co-ordinates along x axis and y co-ordinates along y axis will be like below.