Name: _______________________
Mr. Kepple
Vector Problem Set
2D Kinematics – HW#1
Date: ___________ Period: _____
I. RESOLVING VECTORS
1. Calculate the horizontal and vertical components of each vector. These vectors are expressed using
matrix notation which is just <magnitude, direction>.
(a) ⃑
〈
(b) ⃑
〉
〈
〉
𝑑𝑥
cos
m
𝑣𝑥
cos
− 8 m /s
𝑑𝑦
sin
7 m
𝑣𝑦
sin
m/s
(c) ⃑
〈
(d) ⃑
〉
〈
〉
𝐹𝑥
cos
N
𝑑𝑥
cos
𝐹𝑦
sin
m
𝑑𝑦
sin
m/s
m/s
2. Resolve each vector first by drawing the horizontal and vertical components as referenced to the
dotted line. Next label the magnitude of each component in terms of the given symbols.
𝑟 sin 𝜃
⃑
𝜃
𝑣 sin 𝜃
𝑟 cos 𝜃
𝑣 cos 𝜃
⃑
𝐹 sin 𝜃
𝑝 cos 𝜃
𝑝 sin 𝜃
⃑
⃑
𝑎 sin 𝜃
𝐹 cos 𝜃
⃑
⃑
𝑎 cos 𝜃
𝑑 sin 𝜃
𝑑 cos 𝜃
3. A block slides down a frictionless ramp as shown in the
picture. Even though gravity pulls the block straight down,
the ramp prevents the block from accelerating straight
down. Instead, the block accelerates down the ramp.
(a) Deter ine the
𝑎
agnitude of the block’ acceleration.
𝑔 sin 𝜃
𝜃
𝑔 cos 𝜃
𝜃
(b) Briefly comment on how changing the angle of the
incline affects the acceleration of the block.
As the angle of the incline increases, the acceleration
increases to a maximum value of 9.8 m/s2 when 𝜃
𝑔⃑
𝑔 sin 𝜃
.
II. ADDING VECTORS GRAPHICALLY
⃑⃑
⃑
⃑
⃑
4. You are only drawing pictures for this part, there are no calculations. Use the vectors shown above
to draw the resultant vector from each vector operation. Your picture don’t have to be perfect, but
they should accurately represent both the magnitude and direction of the resultant vector.
⃑⃑
(a) ⃑
(b) ⃑
𝑏⃑⃑
𝑏⃑⃑
𝑎⃑
⃑
(c) ⃑⃑
𝑐⃑
𝑑⃑
𝑎⃑
𝑑⃑
⃑
𝑑⃑
𝑐⃑
(d) ⃑ − ⃑⃑
𝑑⃑
𝑏⃑⃑
(e) ⃑ − ⃑⃑
𝑎⃑ − 𝑏⃑⃑
𝑏⃑⃑
(f) ⃑ − ⃑
𝑑⃑ − 𝑏⃑⃑
𝑑⃑
𝑎⃑
𝑏⃑⃑
𝑎⃑ − 𝑑⃑
𝑑⃑
𝑎⃑
𝑏⃑⃑
II. ADDING VECTORS BY COMPONENTS
5. Suppose that vector ⃑ from above has magnitude 12 meters and direction 30° and vector ⃑⃑ has
magnitude 10 meters and direction 180°. What is the magnitude and direction of ⃑ ⃑⃑ ?
𝑎⃑
𝑏⃑⃑
𝑏⃑⃑
𝑎
𝑏
𝑎⃑
𝑎
𝑏
𝑎
𝑏
𝑎𝑥
cos
4m
𝑎𝑦
sin
m
𝑏𝑥
cos 8
𝑏𝑦
sin 8
−
m
m
𝜃
tan−1
𝜃
tan−1
𝜃
8
𝑎𝑥
𝑏𝑥
4−
m
𝑎𝑦
𝑎𝑥
𝑏𝑦
𝑏𝑥
2
𝑎𝑦
2
𝑏𝑦
2
2
Name: _______________________
Mr. Kepple
Chapter 4: Questions
2D Kinematics – HW#2
Date: ___________ Period: _____
1. You are to launch a rocket, from just above the ground, with one of the following initial conditions:
(1)
(2)
,
(3)
,
In your coordinate system,
,
(4)
runs along level ground and
,
.
increases upward.
(a) Rank the vectors according to the launch speed of the projectile, greatest first. Justify your answer.
All tie. The launch speed of the projectile is the magnitude of the initial velocity vector.
All four vectors above magnitude 20 m/s in the 𝑥-dirction and 70 in the 𝑦-direction,
therefore all four vectors have the magnitude initial velocity, the only difference is in
the direction of the launch.
(b) Rank the vectors according to the time of flight of the projectile, greatest first. Justify your answer.
1 and 2 tie, then 3 and 4. The time of flight is determined by the vertical component of
the initial velocity and rockets 1 and 2 have the same vertical component. Rockets 3
and 4 also have the same vertical component, however they are directed downward,
towards the ground, which means they will hit the ground sooner.
2. The figure shows three paths for a football kicked from ground level. Ignore air resistance and rank
the paths, greatest first, according to the following and justify each of your answers.
(a) maximum height
All tie. In the figure, the maximum height
is represented by the dashed line. Each
football is shown to reach the same
maximum height.
(b) initial vertical velocity components
(c) initial horizontal velocity components
All tie. Since each football reaches the
same maximum height they must have
been launched with the same initial
vertical velocity.
3, 2, 1. The football which travels the
greatest horizontal distance was
launched with the greatest initial
horizontal velocity.
(d) time of flight
(e) initial speed
All tie. The time of flight is determined
by the initial vertical velocity which is the
same for all three footballs.
3, 2, 1. Since the vertical components
are all equal the football with the
greatest horizontal components will
have the greatest magnitude.
3. The figure below shows three situations in which identical projectiles are launched (at the same
level) at identical initial speeds and angles. The projectiles do not land on the same terrain, however.
Rank the situations according to the final speeds of the projectiles just before they land, greatest first.
Justify your answer.
a, b, c. Since each projectile has the same horizontal velocity the one with the greatest
speed will be the one with the greatest vertical velocity. At the maximum height each
projectile has vertical velocity of zero and acceleration of –g. Therefore the projectile
that falls the greatest distance from maximum height has the greatest final speed.
4. An airplane flying horizontally at a constant speed of 350 km/h over level ground releases a bundle
of food supplies. Ignore air resistance and justify all of your answers, what are the bundle’s initial
(a) vertical component of velocity
(b) horizontal component of velocity
The initial vertical component of velocity
is zero since the bundle was launched in
the horizontal direction.
The initial horizontal component of
velocity is 350 km/h since the bundle
was launched in the horizontal direction
its speed will match the airplane’s speed.
(c) What is its horizontal component of velocity
just before hitting the ground?
(d) If the airplane’s speed were, instead, 450
km/h, would the time of fall be longer, shorter,
or the same? Justify your answer.
Since the bundle is only accelerating in
the vertical direction, the horizontal
component of velocity will be a constant
350 km/h for the entire time of flight.
The time of flight would be the same
since it is not affected by the horizontal
component of velocity, only the vertical
component which still starts from zero.
5. A ball is shot from ground level over level ground at a
certain initial speed. The graph to the right gives the range
of the ball versus its launch angle . Rank the three
lettered points on the plot, greatest first, according to the
following and justify all of you answers.
(a) the total flight time of the ball
(b) the ball’s speed at maximum height
c, b, a. The flight time is determined by
the vertical component of the initial
velocity. The ball with the greatest launch
angle has the greatest initial vertical
velocity and greatest flight time.
a, b, c. At maximum height the vertical
velocity is zero so the speed is equal to
the horizontal velocity. The ball with the
lowest launch angle has the greatest
horizontal velocity and greatest speed.
Projectile Motion Problem Set
2D Kinematics – HW#3
Mr. Kepple
Name: _______________________
Date: ___________ Period: _____
1. A tiger leaps horizontally from a 7.5 meter high rock with a speed of 3.2 m/s. How far from the base
of the rock will she land?
1
𝑥
𝑣0𝑦 𝑡 − 𝑔𝑡 2
𝑦 − 𝑦0
2
𝑡
2
(3 2)(1 237)
𝑥
2𝑦0
2(7 5)
𝑔
(9 8)
𝑡
𝑥
1
− 𝑔𝑡 2
− 𝑦0
𝑣0𝑥 𝑡
3 5 m
𝑥≈4 m
1 237 s
2. A diver running 2.3 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches
the water below. (a) How high was the cliff and (b) how far from its base did the diver hit the water?
𝑦 − 𝑦0
1
2
𝑥
1
− 𝑔𝑡 2
− 𝑦0
𝑦0
𝑥
𝑣0𝑦 𝑡 − 𝑔𝑡 2
2
(2 3)(3)
𝑥
(4 )(3)2
𝑣0𝑥 𝑡
6 m
𝑦0 ≈ 44 m
3. In the 1991 World Track and Field Championships in Tokyo, Mike Powell jumped 8.95 m, breaking by
a full 5 cm the 23-year long-jump record set by Bob Beamon. Assume that Powell’s speed on takeoff
was 9.5 m/s (about equal to that of a sprinter) and that
m/s² in Tokyo. How much less was
Powell’s range than the maximum possible range for a particle launched at the same speed?
𝑣0 2 sin(2𝜃)
𝑔
𝑅
𝑅
( 5)2 sin(2 ∗ 45)
𝑅
2
m
𝑑
2
𝑑
−
25 m
𝑑 ≈ 25 cm
5
These last two problems are more challenging and are thus worth double the amount of points.
4. You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to
find its maximum horizontal range. You shoot the gun straight up, and it takes 4.0 s for the dart to land
back at the barrel. What is the maximum horizontal range of your gun?
1
𝑦 − 𝑦0
𝑣0𝑦 𝑡 − 𝑔𝑡 2
−
𝑣0𝑦 𝑡 − 𝑔𝑡 2
𝑅
2
1
2
𝑣0𝑦
2
𝑔𝑡
2
𝑅
𝑔𝑡
2
𝑣0 2 sin(2𝜃)
𝑔
sin(2 ∗ 45)
𝑔
𝑔𝑡 2
4
𝑅
𝑅
𝑔(4)(4)
4
4𝑔 ≈ 3 m
5. A shot-putter throws the shot (mass = 7.3 kg) with an initial speed of 14.4 m/s at a 34.0° angle to the
horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete’s hand at a
height of 2.10 m above the ground.
𝑦 − 𝑦0
1
𝑥
𝑣0𝑦 𝑡 − 𝑔𝑡 2
2
𝑥
1
(𝑣0 sin 𝜃)𝑡 − 𝑔𝑡 2
− 𝑦0
𝑣0𝑥 𝑡
(𝑣0 cos 𝜃)𝑡
2
4 𝑡2 −
𝑡
52𝑡 − 2 1
1 72 s
𝑥
(11 4)(1 72)
𝑥
22 34 m
𝑥 ≈ 22 3 m
Name: _______________________
Mr. Kepple
Projectile Motion FRQ
2D Kinematics – HW#4
Date: ___________ Period: _____
You have been hired by a football team to use physics to improve their kicking game. Before you can
give any recommendations to the coach, you will need to gather some data first. You go out on the
field and watch the kicker make kicks starting from ground level on the flat horizontal field.
30°
52 yards
During one of these kicks, you measure the angle of the kick to be 30° with respect to the ground and
that the football travels a horizontal range of 52 yards.
(a) What was the range of the football in meters? (1 yd = 0.9144 meters)
52 yd ×
0.9144 m
= 47.5488 m ≈ 48 yd
1 yd
(b) Based on this kick, what initial velocity, in meters per second, is the kicker capable of giving to a
football?
𝑣0 2 sin 2𝜃
𝑅=
𝑔
𝑣0 =
𝑅𝑔
sin 2𝜃
𝑣0 =
47.5488 9.8
sin 60
𝑣0 = 23.196 m/s
𝑣0 ≈ 23 m/s
(c) Calculate the maximum possible range, in meters, that the kicker could kick a football.
𝑣0 2 sin 2𝜃
𝑅=
𝑔
𝑅=
23.196
2
sin 2 ∙ 45
9.8
𝑅 = 54.905 m
𝑅 ≈ 55 m
3m
The picture shows a regulation football field with the goal post. The bottom bar of the goal post is 3
meters above the ground. A field goal must clear this height when it reaches the goal post.
Based on the information gathered during practice, you are going to report back to the coach the
maximum field goal range of the team’s kicker. This information is extremely valuable to the football
coach in order to determine what the team should do on 4th down. If the ball is outside the maximum
field goal range of the kicker, then the team will have to punt. However, if they are within the kicker’s
range, the team can score more points by kicking a field goal.
(d) Based on the kicker’s ideal kick that you calculated in part (c), how much time will it take for the
football to reach the goal post?
1
𝑦 − 𝑦0 = 𝑣0𝑦 𝑡 − 𝑔𝑡 2
2
1
𝑦 − 0 = 𝑣0 sin 𝜃 𝑡 − 𝑔𝑡 2
2
3 = 23.196 sin 45 𝑡 − 4.9𝑡 2
4.9𝑡 2 − 16.40𝑡 + 3 = 0
𝑡 = 3.1527 s
𝑡 ≈ 3.2 s
(e) Using the time from part (d), calculate the maximum field goal range of the kicker, in yards. This is
the maximum distance the kicker can be away from the goal post and still clear the bottom bar.
𝑥 = 𝑥0 + 𝑣0 cos 𝜃 𝑡
𝑥 = 0 + 23.196 cos 45 3.1527
𝑥 = 51.711 m
𝑥 = 51.711 m ×
1 yd
= 56.552 yd ≈ 56.6 yd
0.9144 m