College Physics 3.33 – A cannon fired from a battleship


The cannon on a battleship can fire a shell a maximum distance of 32.0 km.

(a) Calculate the initial velocity of the shell.

(b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.)

(c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37×103 km . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?


Solution:

Part A

We are given the range of the projectile motion. The range is 32.0 km. We also know that for the projectile to reach its maximum distance, it should be fired at 45°. So from the formula of range,

\displaystyle \text{R}=\frac{\text{v}_0^2\:\sin 2\theta _0}{\text{g}}

we can say that \sin 2\theta _0=\sin \left(2\times 45^{\circ} \right)=\sin 90^{\circ} =1. So, we have

\displaystyle \text{R}=\frac{\text{v}_0^2}{\text{g}}

We can solve for v0 in terms of the other variables. That is

\displaystyle \text{v}_0=\sqrt{\text{gR}}

Substituting the given values, we have

\displaystyle \text{v}_0=\sqrt{\left(9.81\:\text{m/s}^2\right)\left(32\times 10^3\:\text{m}\right)}

\displaystyle \text{v}_0=560.29\:\text{m/s}        ◀

Part B

We are solving for the maximum height here, which happened at the mid-flight of the projectile. The vertical velocity at this point is zero. Considering all this, the formula for the maximum height is derived to be

\displaystyle \text{h}_{\text{max}}=\frac{\text{v}_{0_y}^2}{2\text{g}}

The initial vertical velocity, v0y, is calculated as

\displaystyle \text{v}_{\text{0y}}=\text{v}_0\sin \theta _0=\left(560.29\:\text{m/s}\right)\sin 45^{\circ} =396.18\:\text{m/s}

Therefore, the maximum height is

\displaystyle \text{h}_{\text{max}}=\frac{\left(396.18\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)}

\displaystyle \text{h}_{\text{max}}=8000\:\text{m}       ◀

Part C

Consider the following figure

A right triangle is formed with the legs, the horizontal distance and the radius of the earth, and the hypotenuse is the sum of the radius of the earth and the distance d, which is the unknown in this problem. Using Pythagorean Theorem, and converting all units to meters, we have

\displaystyle \text{R}^2+\left(32.0\times 10^3\:\text{m}\right)^2=\left(\text{R}+\text{d}\right)^2

\displaystyle \left(6.37\times 10^6\:\text{m}\right)^2+\left(32.0\times 10^3\:\text{m}\right)^2=\left(6.37\times 10^6+\text{d}\right)^2

\displaystyle \text{d}=\sqrt{\left(6.37\times \:10^6\:\right)^2+\left(32.0\times 10^3\:\right)^2}-6.37\times \:10^6

\displaystyle \text{d}=80.37\:\text{m}       ◀

This error is not significant because it is only about 1% of the maximum height computed in Part B.


Farmer wants to Fence off his Four-Sided Plot with missing Side| Vector Addition and Subtraction| Analytical Method| College Physics| Problem 3.22


A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in the figure, and then correctly calculates the length and orientation of the fourth side D. What is his result?

8

SOLUTION:

We know that the sum of the 4 vectors is zero. So, Vector D is calculated as

D=-A-B-C

We solve for the x-component first. The x-component of vector D is

D_x=-A_x-B_x-C_x

D_x=-\left(4.70\:km\right)cos\left(-7.50^{\circ} \right)-\left(2.48\:km\right)cos106^{\circ} -\left(3.02\:km\right)cos161^{\circ}

D_x=-1.12\:km

We solve for the y-component.

D_y=-A_y-B_y-C_y

D_y=-\left(4.70\:km\right)sin\left(-7.50^{\circ} \right)-\left(2.48\:km\right)sin106^{\circ} -\left(3.02\:km\right)sin161^{\circ}

D_y=-2.75\:km

Since we already know the x and y component of vector D, we can finally solve for the distance of vector D.

D=\sqrt{\left(D_x\right)^2+\left(D_y\right)^2}

D=\sqrt{\left(-1.12\:km\right)^2+\left(-2.75\:km\right)^2}

D=2.97\:km

The corresponding direction of vector D is

\theta =tan^{-1}\left|\frac{D_x}{D_y}\right|

\theta =tan^{-1}\left|\frac{1.12\:km}{2.75\:km}\right|

\theta =22.2^{\circ} \:W\:of\:S


College Physics 3.21 – Vector Addition and Subtraction


You fly 32.0 km in a straight line in still air in the direction 35.0° south of west.

(a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.)

(b) Find the distances you would have to fly first in a direction 45.0º south of west and then in a direction 45.0° west of north. These are the components of the displacement along a different set of axes—one rotated 45.0°. 


Solution:

Part A

The component along the south direction is

D_s=Rsin\theta =\left(32.0\:km\right)sin\:35^{\circ} =18.4\:km

The component along the west direction is

D_w=Rcos\theta =\left(32.0\:km\right)cos35^{\circ} =26.2\:km

Part B

Consider the following figure with the rotated axes x’-y’.

7

The component along the southwest direction is

D_{SW}=Rcos\theta '=\left(32.0\:km\right)cos10^{\circ} =31.5\:km

The component along the northwest direction is

D_{NW}=Rsin\theta '=\left(32.0\:km\right)cos10^{\circ} =5.56\:km


College Physics 3.20 – Triangular Piece of Flat Land


A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.61. She then correctly calculates the length and orientation of the third side C. What is her result?

The displacement vectors A, B, and C formed a closed triangle. A is 80 m directed 21° clockwise from the positive x-axis, vector B has a magnitude of 105 m and is directed 11° from the positive y-axis. Find for the magnitude and direction of vector C.
Figure 3.61: The displacement vectors A, B, and C.

Solution:

Since the figure is closed, we know that the sum of the three vectors is equal to 0. So, we have 

\vec{A}+\vec{B}+\vec{C}=0

Solving for C, we have

\vec{C}=-\vec{A}-\vec{B}

Next, we solve for the x and y components of Vectors A and B. 

For Vector A, the x and y components are

A_x=\left(80\:m\right)cos\left(21^{\circ} \right)=74.6864\:m

A_y=-\left(80\:m\right)sin\left(21^{\circ} \right)=-28.6694\:m

For Vector B, the x and y components are

B_x=-\left(105\:m\right)sin\:11^{\circ} =-20.0349\:m

B_y=\left(105\:m\right)cos\:11^{\circ} =103.0709\:m

Solve for the components of Vector C. 

C_x=-A_x-B_x=-74.6864\:m\:-\left(-20.0349\:m\right)=-54.6515\:m

C_y=-A_y-B_y=-\left(-28.6694\:m\right)-\left(103.0709\:m\right)=-74.4015\:m

Therefore, the length of Vector C is

C=\sqrt{\left(C_x\right)^2+\left(C_y\right)^2}

C=\sqrt{\left(-54.6515\:m\right)^2+\left(-74.4015\:m\right)^2}

C=92.3167\:m

The direction of Vector C is

\theta =tan^{-1}\left(\frac{C_y}{C_x}\right)=tan^{-1}\left(\frac{74.4015\:m}{54.6515\:m}\right)=53.7^{\circ} \:\:South\:of\:West


College Physics 3.18 – Driving in a straight line vs east and north distances


You drive 7.50 km in a straight line in a direction 15° east of north.

(a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.)

(b) Show that you still arrive at the same point if the east and north legs are reversed in order.


Solution:

Part A

Consider the following figure:

North and East Components of the given displacement
North and East Components of the given displacement

The east distance is the component in the horizontal direction.

D_E=7.50\:km\:\cdot sin\:\left(15^{\circ} \right)

D_E=1.94\:km

The north distance is the vertical component

D_E=7.50\:km\cdot cos\left(15^{\circ} \right)

D_E=7.24\:km

Part B

3.19

Based from the figure, we can easily see that the order is reversible in the addition of vectors. We say that D_E+D_N=D_N+D_E


College Physics 3.17 – Reverse addition of vectors


Repeat Problem 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B+A=A+B) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking your other path.

Reverse addition of the two vectors A and B.
Reverse addition of the two vectors A and B.

Solution:

Consider the right triangle formed by the legs A, B, and R. We know that A is 18 m, B is 25 m, and we are solving for the magnitude of R. We can do this by using the Pythagorean Theorem. That is

R=\sqrt{A^2+B^2}

R=\sqrt{\left(18\:m\right)^2+\left(25\:m\right)^2}

R=\sqrt{324+625}

R=\sqrt{949}

R=30.8\:m

So, the distance is about 30.8 meters from the starting point. To solve for the value of the unknown angle, φ, we can use the tangent function. That is 

tan\:\phi =\frac{A}{B}

tan\:\phi =\frac{18\:m}{25\:m}

\phi =tan^{-1}\left(\frac{18}{25}\right)

\phi =35.75^{\circ}

Therefore, the compass angle is 35.75^{\circ} \:West\:of\:North


College Physics 3.16 – Resultant of two vectors


Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.60, then this problem asks you to find their sum R = A + B .)

The resultant R and θ are shown in the figure. We are given the vectors A and B, and the right triangle is formed.
Figure 3.60: The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique.


Solution:

Consider the right triangle formed by the legs A, B, and R. We know that A is 18 m, B is 25 m, and we are solving for the magnitude of R. We can do this by using the Pythagorean Theorem. That is

R=\sqrt{A^2+B^2}

R=\sqrt{\left(18\:m\right)^2+\left(25\:m\right)^2}

R=\sqrt{324+625}

R=\sqrt{949}

R=30.8\:m

So, the distance is about 30.8 meters from the starting point. To solve for the value of the unknown angle, θ, we can use the tangent function. That is 

tan\:\theta =\frac{B}{A}

tan\:\theta =\frac{25\:m}{18\:m}

\theta =tan^{-1}\left(\frac{25}{18}\right)

\theta =54.25^{\circ}

So, the compass reading, can be solved by taking the complimentary angle, \phi as shown in the figure.

The angle 𝜙 is the complimentary angle of 𝜃.
The angle ɸ is the complimentary angle of θ.

\phi =90^{\circ} -54.25^{\circ}

\phi =35.75^{\circ}

Therefore, the compass angle is 35.75^{\circ} \:West\:of\:North


College Physics 3.14 – Distance and displacement of a path


Find the following for path D in Figure 3.58:

(a) the total distance traveled and

(b) the magnitude and direction of the displacement from start to finish.

In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

3.1
Figure 3.58 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Solution:

Part A

The total distance traveled following Path D is 

d=\left(2\times 120\:m\right)+\left(6\times 120\:m\right)+\left(4\times 120\:m\right)+\left(1\times 120\:m\right)

d=1560\:m

Part B

Treat all motions upward and to the right positive, while downward and to the left negative. 

The displacement in the x-direction is

s_x=0+720\:m+0-120\:m=600\:m

The displacement in the y-direction is

s_y=-240\:m+0+480\:m+0=240\:m

The total displacement is

s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}=\sqrt{\left(600\:m\right)^2+\left(240\:m\right)^2}=646.22\:m

The direction angle from th x-axis is given by

\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|=tan^{-1}\left|\frac{240}{600}\right|=21.8^{\circ} \:North\:of\:East

Therefore, the displacement is 646.22 m, directed at 21.8 degrees north of east.


College Physics 3.13 – Distance and Displacement


Find the following for path C in Figure 3.58:

(a) the total distance traveled and

(b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

The figure shows C starts from a point, then goes up 1 block, then 5 blocks to the right, then 2 blocks downward, 1 block to the left, 1 block upward, then 3 blocks to the left
Figure 3.58 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Solution:

Part A

The total distance traveled following Path C is 

d=\left(1\times 120\:m\right)+\left(5\times 120\:m\right)+\left(2\times 120\:m\right)+\left(1\times 120\:m\right)+\left(1\times 120\:m\right)+\left(3\times 120\:m\right)

d=1560\:m

Part B

Treat all motions upward and to the right positive, while downward and to the left negative. 

The displacement in the x-direction is

s_x=0+600+0-120+0-360=120\:m

The displacement in the y-direction is

s_y=120+0-240+0+120+0=0\:m

The total displacement is

s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}

s=\sqrt{\left(120\:m\right)^2+\left(0\:m\right)^2}

s=120\:m

The direction angle from th x-axis is given by

\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|

\theta _x=tan^{-1}\left|\frac{0\:m}{120\:m}\right|

\theta _x=0^{\circ}

Therefore, the displacement is 120 m, east.