Vector Problem Set 2D Kinematics – HW#1 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) ⟨ ⟩

(c) ⟨ ⟩ (d) ⟨ ⟩

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.

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 agnitude of the block’ acceleration. (b) Briefly comment on how changing the angle of the incline affects the acceleration of the block.

Name: _______________________

Date: ___________ Period: _____

𝜃

𝜃

𝜃

��

𝑑𝑦 sin 7 m

𝑑𝑥 cos m 𝑣𝑥 cos − 8 m /s

𝑣𝑦 sin m/s

𝐹𝑦 sin m

𝐹𝑥 cos N 𝑑𝑥 cos m/s

𝑑𝑦 sin m/s

𝑣 cos𝜃

𝑣 sin𝜃 𝑟 cos𝜃

𝑟 sin 𝜃 𝑝 cos𝜃

𝑝 sin𝜃

𝐹 cos𝜃

𝐹 sin𝜃

𝑎 sin𝜃

𝑎 cos𝜃 𝑑 cos𝜃 𝑑 sin𝜃

𝑔 cos𝜃

𝑔 sin𝜃

𝑎 𝑔 sin𝜃

As the angle of the incline increases, the acceleration increases to a maximum value of 9.8 m/s2 when 𝜃 .

Mr. Kepple

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) (c)

(d) − (e) − (f) − 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 ?

𝑐 𝑑

𝑐

𝑑

��

��

�� ��

�� 𝑑

��

𝑑

��

��

�� − �� 𝑑 − ��

��

𝑑

��

𝑑

�� − 𝑑

𝑎𝑦 sin m

𝑎𝑥 cos 4 m

𝑏𝑦 sin 8 m

𝑏𝑥 cos 8 − m

��

��

�� ��

𝑎 𝑏 𝑎𝑥 𝑏𝑥 2 𝑎𝑦 𝑏𝑦

2

𝑎 𝑏 4 − 2 2

𝑎 𝑏 m

𝜃 tan−1 𝑎𝑦 𝑏𝑦𝑎𝑥 𝑏𝑥

𝜃 tan−1

𝜃 8

Chapter 4: Questions 2D Kinematics – HW#2

1. You are to launch a rocket, from just above the ground, with one of the following initial conditions:

(1) ,

(2) ,

(3) ,

(4) , .

In your coordinate system, runs along level ground and increases upward.

(a) Rank the vectors according to the launch speed of the projectile, greatest first. Justify your answer.

(b) Rank the vectors according to the time of flight of the projectile, greatest first. Justify your answer.

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

(b) initial vertical velocity components (c) initial horizontal velocity components

(d) time of flight (e) initial speed

Name: _______________________

Date: ___________ Period: _____

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.

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.

All tie. In the figure, the maximum height

is represented by the dashed line. Each

football is shown to reach the same

maximum height.

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.

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.

Mr. Kepple

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.

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

(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.

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

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.

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.

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.

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 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?

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? 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?

Name: _______________________

Date: ___________ Period: _____

𝑡 1 237 s

𝑦 − 𝑦0 𝑣0𝑦𝑡 −1

2𝑔𝑡2

− 𝑦0 −1

2𝑔𝑡2

𝑡 2𝑦0

𝑔

2(7 5)

(9 8)

𝑥 (3 2)(1 237)

𝑥 3 5 m

𝑥 ≈ 4 m

𝑥 𝑣0𝑥𝑡

𝑦0 (4 )(3)2

𝑦0 ≈ 44 m

𝑦 − 𝑦0 𝑣0𝑦𝑡 −1

2𝑔𝑡2

− 𝑦0 −1

2𝑔𝑡2

𝑥 (2 3)(3)

𝑥 6 m

𝑥 𝑣0𝑥𝑡

𝑅 𝑣02 sin(2𝜃)

𝑔

𝑅 ( 5)2 sin(2 ∗ 45)

𝑅 2 m

𝑑 2 − 5

𝑑 25 m

𝑑 ≈ 25 cm

Mr. Kepple

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?

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.

𝑅 𝑣02 sin(2𝜃)

𝑔

𝑅 𝑔𝑡

2 2 sin(2 ∗ 45)

𝑔

𝑅 𝑔𝑡2

4 𝑔(4)(4)

4

𝑅 4𝑔 ≈ 3 m

𝑣0𝑦 𝑔𝑡

2

𝑦 − 𝑦0 𝑣0𝑦𝑡 −1

2𝑔𝑡2

− 𝑣0𝑦𝑡 −1

2𝑔𝑡2

4 𝑡2 − 52𝑡 − 2 1

𝑡 1 72 s

𝑦 − 𝑦0 𝑣0𝑦𝑡 −1

2𝑔𝑡2

− 𝑦0 (𝑣0 sin𝜃)𝑡 −1

2𝑔𝑡2

𝑥 (𝑣0 cos𝜃)𝑡

𝑥 (11 4)(1 72)

𝑥 22 34 m

𝑥 ≈ 22 3 m

𝑥 𝑣0𝑥𝑡

Projectile Motion FRQ 2D Kinematics – HW#4 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. 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) (b) Based on this kick, what initial velocity, in meters per second, is the kicker capable of giving to a football? (c) Calculate the maximum possible range, in meters, that the kicker could kick a football.

Name: _______________________

Date: ___________ Period: _____

30°

52 yards

Mr. Kepple

𝑅 =𝑣0

2 sin 2𝜃

𝑔

𝑣0 = 𝑅𝑔

sin 2𝜃

52 yd ×0.9144 m

1 yd= 47.5488 m ≈ 48 yd

𝑣0 = 47.5488 9.8

sin 60

𝑣0 = 23.196 m/s

𝑣0 ≈ 23 m/s

𝑅 =𝑣0

2 sin 2𝜃

𝑔 𝑅 =

23.196 2 sin 2 ∙ 45

9.8

𝑅 = 54.905 m

𝑅 ≈ 55 m

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? (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.

3 m

3 = 23.196 sin 45 𝑡 − 4.9𝑡2

4.9𝑡2 − 16.40𝑡 + 3 = 0

𝑡 = 3.1527 s

𝑡 ≈ 3.2 s

𝑦 − 𝑦0 = 𝑣0𝑦𝑡 −1

2𝑔𝑡2

𝑦 − 0 = 𝑣0 sin𝜃 𝑡 −1

2𝑔𝑡2

𝑥 = 𝑥0 + 𝑣0 cos𝜃 𝑡

𝑥 = 0 + 23.196 cos 45 3.1527

𝑥 = 51.711 m

𝑥 = 51.711 m ×1 yd

0.9144 m= 56.552 yd ≈ 56.6 yd