Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 3: Two-Dimensional Kinematics

34 problems · 10 figures

3-1 §3.2 source ↗

Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.

Find the following for path A in Figure 3.49: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.

Figure 3.49 in the book ↗
The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Answer

(a) 480 m

(b) 379 m, 18.4º east of north

3-3 §3.2 source ↗

Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.

Find the north and east components of the displacement for the hikers shown in Figure 3.47.

Figure 3.47 in the book ↗

Answer

north component 3.21 km, east component 3.83 km

3-5 §3.2 source ↗

Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.

Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. 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 3.51, then this problem finds their sum R = A + B.)

Figure 3.51 in the book ↗

Answer

19.5 m, 4.65º south of west

3-7 §3.2 source ↗

Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.

(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east, (which is equivalent to subtracting B from A —that is, finding R′=A–B). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º east of south (which is equivalent to subtracting A from B —that is, to finding R′′= B-A=-R′). Show that this is the case.

Answer

(a) 26.6 m, 65.1º north of east

(b) 26.6 m, 65.1º south of west

3-9 §3.2 source ↗

Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.

Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.23.

Figure 3.23 in the book ↗
Example 3.2 Subtracting Vectors Graphically: A Woman Sailing a Boat in the book ↗

Answer

52.9 m, 90.1º with respect to the x-axis.

3-11 §3.2 source ↗

Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.

Find the components of vtot along the x- and y-axes in Figure 3.52.

Figure 3.52 in the book ↗
The two velocities vA and vB add to give a total vtot.

Answer

x-component 4.41 m/s

y-component 5.07 m/s

3-13 §3.3 source ↗

Find the following for path C in Figure 3.53: (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.
Figure 3.53 in the book ↗
The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Answer

(a) 1.56 km

(b) 120 m east

3-15 §3.3 source ↗

Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.54.
Figure 3.54 in the book ↗

Answer

North-component 87.0 km, east-component 87.0 km

3-17 §3.3 source ↗

Repeat Exercise 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.

Answer

30.8 m, 35.8 west of north

3-19 §3.3 source ↗

Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B from A —that is, finding R′=A – B) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B —that is, to find A=B+C. Is that consistent with your result?)

Answer

(a) 30.8 m, 54.2º south of west

(b) 30.8 m, 54.2º north of east

3-21 §3.3 source ↗

You fly 32.0 km in a straight line in still air in the direction 35.0º 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 45.0º west of north. These are the components of the displacement along a different set of axes—one rotated 45º.

Answer

18.4 km south, then 26.2 km west(b) 31.5 km at 45.0º south of west, then 5.56 km at 45.0º west of north

3-23 §3.3 source ↗

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30 km 25.0º south of west; then 5.10 km straight east; then 1.70 km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?

Answer

7.34 km, 63.5º south of east

3-25 §3.4 source ↗

A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

Answer

x = 1.30 m×102 y = 30.9 m.

3-27 §3.4 source ↗

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Answer

(a) 3.50 s

(b) 28.6 m/s (c) 34.3 m/s

(d) 44.7 m/s, 50.2º below horizontal

3-29 §3.4 source ↗

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

Answer

(a) 18.4º

(b) The arrow will go over the branch.

3-31 §3.4 source ↗

Verify the ranges for the projectiles in Figure 3.38(a) for θ=45º and the given initial velocities.
Figure 3.38 in the book ↗
Trajectories of projectiles on level ground. (a) The greater the initial speed v0, the greater the range for a given initial angle. (b) The effect of initial angle θ0 on the range of a projectile with a given initial speed. Note that the range is the same for 15º and 75º, although the maximum heights of those paths are different.

Answer

R=v02 sin 2θ 0 g Forθ=45º, R= v02g

R=91.8 m for v0=30 m/s; R=163 m for v0=40 m/s; R=255m for v0=50 m/s.

3-33 §3.4 source ↗

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?

Answer

(a) 560 m/s

(b) 8.00 × 103 m

(c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

3-35 §3.4 source ↗

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g. How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Answer

1.50 m, assuming launch angle of 45º

3-37 §3.4 source ↗

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The base line is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose service line is 6.40 m from the net?

Answer

θ = 6.1º

yes, the ball lands at 5.3 m from the net

3-39 §3.4 source ↗

Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

Answer

(a) −0.486 m

(b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.

3-41 §3.4 source ↗

An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0º below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

Answer

4.23 m. No, the owl is not lucky; he misses the nest.

3-43 §3.4 source ↗

Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

Answer

No, the maximum range (neglecting air resistance) is about 92 m.

3-45 §3.4 source ↗

In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0º above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45º when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38º will give a longer range than 45º in the shot put.)

Answer

15.0 m/s

3-47 §3.4 source ↗

A football player punts the ball at a 45.0º angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?

Answer

(a) 24.2 m/s

(b) The ball travels a total of 57.4 m with the brief gust of wind.

3-49 §3.4 source ↗

Derive R= v02sin2θ0g for the range of a projectile on level ground by finding the time t at which y becomes zero and substituting this value of t into the expression for x−x0, noting that R=x−x0

Answer

y− y0= 0= v0yt− 12gt2= (v0sinθ)t− 12gt2,

so that t= 2(v0 sin θ)g

x− x0= v0xt= (v0cosθ)t= R, and substituting for t gives:

R= v 0 cosθ 2 v 0 sin θ g = 2 v 0 2 sin θ cos θ g

since 2sinθcosθ= sin2θ, the range is:

R= v 0 2 sin2θg.

3-52 §3.5 source ↗

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45º south of east. What was his total displacement? (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air? (c) What was his total displacement relative to the air mass?

Answer

(a) 35.8 km, 45º south of east

(b) 5.53 m/s, 45º south of east

(c) 56.1 km, 45º south of east

3-54 §3.5 source ↗

Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of 3.50 m/s, and the second a velocity of 4.20 m/s. (a) What is the velocity of the second runner relative to the first? (b) If the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity? (c) What distance ahead will the winner be when she crosses the finish line?

Answer

(a) 0.70 m/s faster

(b) Second runner wins

(c) 4.17 m

3-56 §3.5 source ↗

A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of 25.0º relative to the ground and is caught at the same height as it is released. What is the initial velocity of the ball relative to the quarterback ?

Answer

17.0 m/s, 22.1º

3-58 §3.5 source ↗

(a) A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a direction 5.0º south of west. It is in the jet stream, which is blowing at 35.0 m/s in a direction 15º south of east. What is the velocity of the airplane relative to the Earth? (b) Discuss whether your answers are consistent with your expectations for the effect of the wind on the plane’s path.

Answer

(a) 230 m/s, 8.0º south of west

(b) The wind should make the plane travel slower and more to the south, which is what was calculated.

3-60 §3.5 source ↗

(a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 20º south of east (as in Exercise 3.58). Its direction of motion relative to the Earth is 45.0º south of west, while its direction of travel relative to the air is 5.00º south of west. What is the airplane’s speed relative to the air mass? (b) What is the airplane’s speed relative to the Earth?

Answer

(a) 63.5 m/s

(b) 29.6 m/s

3-62 §3.5 source ↗

The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of 2.20 m/s in a direction 30.0º east of north relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction of 50.0º south of west relative to the Earth. What is the velocity of the wind relative to the water?

Answer

6.68 m/s, 53.3º south of west

3-64 §3.5 source ↗

(a) Use the distance and velocity data in Figure 3.59 to find the rate of expansion as a function of distance.

(b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant for universal expansion and the time back to the Big Bang are determined, respectively.

Figure 3.59 in the book ↗
Five galaxies on a straight line, showing their distances and velocities relative to the Milky Way (MW) Galaxy. The distances are in millions of light years (Mly), where a light year is the distance light travels in one year. The velocities are nearly proportional to the distances. The sizes of the galaxies are greatly exaggerated; an average galaxy is about 0.1 Mly across.

Answer

(a) H average = 14.9 km/s Mly

(b) 20.2 billion years

3-66 §3.5 source ↗

A ship sailing in the Gulf Stream is heading 25.0º west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is 4.80 m/s 5.00º west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the United States.)

Answer

1.72 m/s, 42.3º north of east

3-71 §3.5 source ↗

Critical Thinking Two identical spring-loaded launchers are fixed to the end of a table. Launcher A is inclined 40º above the horizontal and launcher B is inclined 50º above the horizontal. (a) Which of these launchers will fire the ball the greatest vertical height above the launcher before it crashes into the ground? (b) Based on experimental data, an equation that fits the data is suggested for the vertical height above the launcher, y, which may not be correct: y=K V 0 2 sinθ , where K is a constant with appropriate units, V 0 is the initial velocity of the launcher, and θ is the angle relative to the x-axis of the launcher. Is this equation consistent with your answer from part a? Explain why or why not. Does this equation make sense? Explain why or why not. (c) Graph the magnitude of the velocity vs. vertical height for both launcher A and B.

Answer

(a) Since sin 2 50> sin 2 40, then B reaches the greatest height.

(b) i. Yes, it is consistent because sin50> sin40. ii. No, it does not make sense because y is proportional to sin 2 θ .

(c)

The figure shows speed (absolute value of v) graphed as a function of position y. The plot of A is linear, with a negative slope, starting at v sub zero A and ending at speed zero. The plot of B is linear, with the same negative slope as that of A, starting at v sub zero B, which is greater than v sub zero A, and ending at speed zero.

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Problems and answers are adapted from College Physics 2e by Paul Peter Urone, Roger Hinrichs, Kim Dirks and Manjula Sharma, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.