Chapter 3 Solutions
Answers to Questions
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Q13
Yes. Upon being dropped, the paper goes from zero velocity to some velocity downward.
No. The motion is one of constant velocity since the ball moves equal distances in equal times.
Yes. A has positive acceleration (speeding up).
B has negative acceleration (slowing down).
Both will fall with the same acceleration.
The crumpled one will be less affected by air resistance and will reach the floor first.
Because there is no air resistance in the evacuated tube, both the crumpled and the uncrumpled sheets of paper
will reach the bottom of the tube at the same moment.
In the absence of air friction, Aristotle was wrong. In the presence of air friction, the shape and mass of the object
can both affect the acceleration of a falling object. It is possible for a heavy object to fall slower than a light object
and vice versa.
The distance traveled in a 0.10 s interval just before it hits the water will be greater than just after it is dropped
because the velocity will have increased.
No. Constant acceleration would require the v vs. t curve to be a straight line. It shows curvature for the same time.
(Acceleration is increasing.)
a. The thrown ball will reach the ground first. This is because its increase in velocity due to gravity though equal
to that of the dropped ball is in addition to its initial velocity. The initial velocity continues to move the ball downward
at a constant rate which adds to the amount of distance traversed because of the acceleration.
b. No. The only acceleration is that due to gravity which is the same for both balls.
The direction of the velocity vector during this time is upward. The direction of the acceleration vector is downward.
Acceleration does not decrease. The decrease in the magnitude of the velocity is a direct result of the constant,
Earthward directed acceleration.
Yes. The velocity of the rock was decreasing as it ascended. Thus more time was needed to traverse an equal
distance when it was traveling more slowly.
The velocity changes direction at the apex (top) of the trajectory.
velocity
time
Q14
Q15
Q16
Q17
Q18
Q19
Q20
Q21
Q22
No, acceleration is due to gravity that has not gone away!
No. Acceleration is in the downward direction throughout the motion.
The acceleration will be constant (and never zero) because it is due to gravity. The velocity will not be constant
decreasing on the way up, reaching zero then increasing as the ball rolls back down the incline.
Both reach the floor at the same time. They both have the same initial zero y-component of velocity, undergo the
same acceleration, and drop the same vertical distance. The horizontal part of the motion is immaterial.
The rolled ball has the larger velocity. In addition to the vertical component (which is the same as that of the
dropped ball), it has a horizontal component that the dropped ball lacks.
Of course. Horizontal velocity is not influenced by vertical velocity. For example, a kicked ball has both horizontal
and vertical velocity. The only change in velocity comes from acceleration due to gravity.
Yes. Although the horizontal component remains constant, the vertical component increases downward. This
changes the resultant vector that is determined by both magnitude as well as the vertical component of direction.
(See figures below.)
No. Horizontal velocity has no influence on the vertical motion.
Above. The sights have been adjusted to compensate for the bullet's drop at great distance. It will be an over
compensation for firing at a near target.
1
Q23
Q24
Q25
Q26
Q27
Q28
Q29
Q30
Q31
Q32
The higher trajectory takes longer. The time of flight is determined by the initial vertical velocity component which
also determines the maximum height reached.
As there is always a horizontal velocity component not equal to zero, there is no time when the velocity of the ball is
equal to zero for either of the shown trajectories.
The ball with the higher velocity is more likely to have been thrown by the fielder with the stronger arm.
No. The higher angled shot spends its time slowing its higher ascent until it stops and descends toward the ground
No. Even though its horizontal velocity component is greater than that of the other shot, it falls sooner to the ground.
A. The shot in direction C will not clear the chair; in order for the shot in direction B to clear the chair it will have to
have a fairly high initial velocity that would likely cause it to overshoot the basket.
Insufficient magnitude will result in the ball falling short of its goal (regardless of direction). Too great a magnitude
will result in the well-aimed shot overshooting the basket.
The arched shot has the advantage of giving the shooter more leeway in getting the shot through the hoop—a wider
range of trajectories that will make it.
The likelihood of undershooting the basket becomes more likely with distance from the basket.
A hard low trajectory throw will reach the receiver in a shorter time interval and thus more unlikely to be intercepted.
However, this type of throw places a greater demand on the requirement that the receiver accurately judges the
moment and place that they and ball will arrive at the catching point.
Answers to Exercises
E1
E2
E3
E4
E5
E6
E7
E8
E9
E10
E11
E12
E13
E14
E15
E16
a. 8 m/s
b. 16 m/s
a. 3.2 m
b. 12.8 m
50 m/s = 112 MPH
a. 0.75 s
b. 2.76 m
22 m/s
30 m/s
a. +5 m/s up
b. - 5 m/s down
a. 10 m
b. 10 m
1.5 s
a. 6 m/s
b. 6 s
a. 0.1667 s
b. 13.9 cm
2.4 m
1.25 m
a. - 6 m/s
b. 5 m/s
a. 1.0 s
b. 6 m
a. 3 s
b. 90 m
Answers to Synthesis Problems
SP1
SP2
SP3
SP4
a.
b.
c.
d.
e.
a.
b.
c.
a.
b.
c.
0
1.6 s
12.8 m
12 m
The ball is moving down.
vA = 15 m/s
vB = 27 m/s
yA = 11.25 m
yB = 29.25 m
No. Each has the same acceleration, hence, the rate of change of their velocity is the same.
0.4 s
XA = 1.2 m; XB = 2.0 m
No. Ball B will reach the edge of the tabletop first and will begin its downward fall before Ball A. Since they
each have the same downward acceleration, Ball A cannot catch Ball B.
a. t30 = 40 s
b. x = 13,840 m
c. t60 = 69.2 s
x = 13,840 m; Same as for the first angle.
2
SP5
SP6
a.
b.
c.
d.
40.2 m/s
18.3 m
0.455 s
1.03 m
a. 0. 65 seconds ( t 
x
4.55m

 0.65s )
vx
7m / s
b. 2.113 m ( y  v oy t 
1 2
at
2
v oy  0
y


1
2
10m / s 2  0.65s   2.113m )
2
c. -0.1625m = -16.25 cm (2.113m-1.95m= 0.1625m), she misses!
3