MR. SURRETTE VAN NUYS HIGH SCHOOL
CHAPTER 2: VECTORS AND PROJECTIES
WORKSHEET SOLUTIONS
Questions 1 - 3 (See classroom board)
4. A string attached to an airborne kite is maintained at an angle of 35 degrees with the horizontal. If a total of 85 m of string is reeled in while bringing the kite back to the ground, what is the horizontal displacement of the kite in the process (assume the kite string doesn’t sag)?
4A.
(1) cos 35o = (adjacent/hypotenuse)
(2) (cos 35o)(hypotenuse) = (adjacent)
(3) (0.8192)(85 m) = 69.6 m
5. A jet airliner moving at 600 mph due east moves into a region where the wind is blowing at 100 mph in a direction 30 degrees north of east. What is the new velocity and direction of the aircraft?
5A.
(1) Define jet vector as S and wind vector as T
(2) Resolve into x and y components:
Sx = 600 Sy = 0
Tx = 100 (cos 30o) Ty = 100 (sin 30o)
Tx = 86.6 Ty = 50
(3) Combine x and y components into a resultant vector. Call the resultant vector R.
Rx = (600 + 86.6) = 686.6
Ry = (0 + 50) = 50
(4) Determine the magnitude of R:
R2 = Rx2 + Ry2
R = [(Rx2 + Ry2)]1/2
R = [(686.6)2 + (50)2]1/2
R = 688.4 mph
(5) Determine the direction of the resultant R vector:
q = arctan (Ry/Rx)
q = arctan (50 / 686.6)
q = 4.2o
(6) Final Answer: 688.4 mph, 4.2o north of east
6. A river flows due east at 2 m/s. A boat crosses the 300 m wide river by maintaining a constant velocity of 10 m/s due north relative to the water. If no correction is made for the current, how far downstream does the boat move by the time it reaches the far shore?
6A.
One way to approach this problem is to determine how much time is spent getting across to the opposite shore:
(1) (300 meters) / (10 meters / second)
(2) Time = 30 seconds
Since the river flows east at 2 m/s:
(3) (30 seconds) (2 meters / second)
(4) Distance = 60 m
7. A girl hikes 4 miles east from A to B and then 3 miles at a heading of 37o degrees north of east from B to C, as shown below. What is the magnitude and direction of the resultant?
7A. (part a)
(1) cos 37o = (adjacent/hypotenuse)
(2) (cos 37o)(hypotenuse) = (adjacent)
(3) (0.799)(3 m) = 2.4 m
7A. (part b)
(1) sin 37o = (opposite/hypotenuse)
(2) (sin 37o)(hypotenuse) = (opposite)
(3) (0.602)(3 m) = 1.8 m
7A. (part c)
(1) Define AB as S and BC as T
(2) Resolve into x and y components:
Sx = 4 Sy = 0
Tx = 3 (cos 37o) Ty = 3 (sin 37o)
Tx = 2.40 Ty = 1.81
(3) Combine x and y components into a resultant vector. Call the resultant vector R.
Rx = (4 + 2.40) = 6.40
Ry = (0 + 1.81) = 1.81
7A. (continued…)
(4) Determine the magnitude of R:
R2 = Rx2 + Ry2
R = [(Rx2 + Ry2)]1/2
R = [(6.40)2 + (1.81)2]1/2
R = 6.6 miles
(5) Determine the direction of the resultant R vector:
q = arctan (Ry/Rx)
q = arctan (1.81 / 6.40)
q = 16o
(6) Final Answer: 6.6 miles, 16o north of east
A BODY PROJECTED HORIZONTALLY
Example 1. A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 17.0 m/s. After 2.5 seconds, find the motorcycle’s: (a) position, (b) distance from the edge of the cliff, and (c) velocity.
(a) POSITION
(1) V = d / t
(2) Vx = D x / D t
(3) D x = Vx D t
(4) D x = (17.0 m/s)(2.5 s)
(5) D x = 42.5 m
(6) D y = Voyt – ½ gt2
(7) D y = - ½ gt2 = - ½ (9.8 m/s2)(2.5 s)2
(8) D y = - 30.6 m
(9) P(x,y) = (42.5 m, - 30.6 m)
(b) DISTANCE FROM CLIFF
(1) Let r = distance
(2) r = [(42.5 m)2 + (- 30.6 m)2] 1/2
(3) r = 52.4 meters
(c ) VELOCITY
(1) Vx = 17 m/s
(2) Vy = -gt = - (9.8 m/s2)(2.5 s) = - 24.5 m/s
(3) V = [(Vx)2 + (Vy)2] ½
(4) V = [(17.0 m/s)2 + (- 24.5 m/s)2] ½
(5) V = 29.8 m/s
(6) q = arctan (Vy / Vx)
(7) q = arctan (- 24.5 / 17.0)
(8) q = - 55.2o
(9) Velocity: 29.8 m/s, 55.2o south of the x axis
HEIGHT AND RANGE OF A BASEBALL
Example 2.
A batter hits a baseball so that it leaves the bat with an initial speed vo = 41.2 m/s at an initial angle
q = 50.4o.
When t = 2.75 seconds find:
(a) the position of the ball at 2.75 seconds
(b) the magnitude and direction of its velocity at 2.75 seconds
(c ) the time when the ball reaches its highest point
(d) the height h when the ball reaches its highest point
(e) the horizontal range R when the ball hits the ground
(a) POSITION
(1) The ball is probably struck a meter or so above the ground, but we neglect this distance and assume that it starts at ground level.
(2) Vx = Vocosq
(3) Vx = (41.2 m/s)(cos 50.4o)
(4) Vx = 26.3 m/s
(5) Vy = Vosinq
(6) Vy = (41.2 m/s)(sin 50.4o)
(7) Vy = 31.7 m/s
(8) Vx = D x / D t
(9) D x = (Vx)(D t)
(10) D x = (26.3 m/s)(2.75 s)
(11) D x = 72.2 m
(12) D y = Vy t - ½ gt2
(13) D y = (31.7 m/s)(2.75 s) – ½ (9.8 m/s2)(2.75 s)2
(14) D y = 50.2 m
(15) P(x,y) = (72.2 m, 50.2 m)
(b) VELOCITY
(1) Vx = 26.3 m/s
(2) Vy = Voy – gt
(3) Vy = 31.7 m/s – (9.8 m/s2)(2.75 s)
(4) Vy = 4.75 m/s
(5) V = [(Vx)2 + (Vy)2] ½
(6) V=[(26.3 m/s)2 + (4.75 m/s)2] ½ = 26.7 m/s
(7) q = arctan (Vy / Vx)
(8) q = arctan ( 4.75 m/s / 26.3 m/s) = 10.3o
(9) Velocity: 26.7 m/s, 10.3o north of x-axis
(c ) TIME TO HIGHEST POINT
(1) Vy = Voy - gt
(2) 0 = Voy – gt
(3) Voy = gt
(4) t = Voy / g
(5) t = (31.7 m/s) / (9.8 m/s2)
(6) t = 3.24 s
(d) MAXIMUM HEIGHT
(1) D y = Voy t- ½ gt2
(2) D y = (31.7 m/s)(3.24 s) – ½ (9.8 m/s2)(3.24 s)2
(3) D y = 51.4 m
(e) RANGE
(1) Vx = D x / D t
(2) D x = (Vx)(D t)
(3) D x = (26.3 m/s)(6.48 s)
(4) D x = 170 m
CHAPTER 2: VECTORS AND PROJECTILES
QUIZ SOLUTIONS
1. A city jogger runs six blocks due east, ten blocks due south, and another two blocks due east. Assume all blocks are of equal size. What is the magnitude of the jogger’s displacement, start to finish?
1A.
x-direction: 8 east = 8
y-direction: 10 south = - 10
displacement: c2 = a2 + b2
c = [(a2 + b2)]1/2
c = [(8)2 + (-10)2]1/2
c = 12.8 blocks
2. A string attached to an airborne kite is maintained at an angle of 42 degrees with the horizontal. If a total of 95 m of string is reeled in while bringing the kite back to the ground, what is the horizontal displacement of the kite in the process (assume the kite string doesn’t sag)?
2A.
(1) cos 42o = (adjacent/hypotenuse)
(2) (cos 42o)(hypotenuse) = (adjacent)
(3) (0.743)(95 m) = 70.6 m
3. A jet airliner moving at 620 mph due east moves into a region where the wind is blowing at 30 mph in a direction 30 degrees north of east. What is the new velocity and direction of the aircraft?
3A.
(1) Define jet vector as S and wind vector as T
(2) Resolve into x and y components:
Sx = 620 Sy = 0
Tx = 30 (cos 30o) Ty = 30 (sin 30o)
Tx = 25.98 Ty = 15
(3) Combine x and y components into a resultant vector. Call the resultant vector R.
Rx = (620 + 25.98) = 646.0
Ry = (0 + 15) = 15
(4) Determine the magnitude of R:
R2 = Rx2 + Ry2
R = [(Rx2 + Ry2)]1/2
R = [(646.0)2 + (15)2]1/2
R = 646.2 mph
(5) Determine the direction of the resultant R vector:
q = arctan (Ry/Rx)
q = arctan (15 / 646.2)
q = 1.33o
(6) Final Answer: 646.2 mph, 1.3o north of east
4. A river flows due east at 3 m/s. A boat crosses the 250 m wide river by maintaining a constant velocity of 8 m/s due north relative to the water. If no correction is made for the current, how far downstream does the boat move by the time it reaches the far shore?
4A.
One way to approach this problem is to determine how much time is spent getting across to the opposite shore:
(1) (250 meters) / (8 meters / second)
(2) Time = 31.25 seconds
Since the river flows east at 3 m/s:
(3) (31.25 seconds) (3 meters / second)
(4) Distance = 93.8 m
5. A girl hikes 4 miles east from A to B and then 7 miles at a heading of 43o degrees north of east from B to C, as shown below. What is the magnitude and direction of the resultant?
Question 5.
5A.
(1) Define A à B as S and A à C as T
(2) Resolve into x and y components:
Sx = 4 Sy = 0
Tx = 7 (cos 43o) Ty = 7 (sin 43o)
Tx = 5.12 Ty = 4.77
(3) Combine x and y components into a resultant vector. Call the resultant vector R.
Rx = (4 + 5.12) = 9.12
Ry = (0 + 4.77) = 4.77
(4) Determine the magnitude of R:
R2 = Rx2 + Ry2
R = [(Rx2 + Ry2)]1/2
R = [(9.12)2 + (4.77)2]1/2
R = 10.3 miles
5A. (continued...)
(5) Determine the direction of the resultant R vector:
q = arctan (Ry/Rx)
q = arctan (4.77 / 9.12)
q = 27.6o
(6) Final Answer: 10.3 miles, 27.6o north of east
6. A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 12.0 m/s.
6a. Find the motorcycle’s position after 0.60 seconds.
A.
(1) Vx = D x / D t
(2) D x = Vx D t
(3) D x = Vx D t
(4) D x = (12.0 m/s)(0.60 s)
(5) D x = 7.2 m
(6) D y = Voyt – ½ gt2 = 0 – ½ gt2
(7) D y = - ½ gt2 = - ½ (9.8 m/s2)(0.60 s)2
(8) D y = - 1.76 m
(9) P(x,y) = (7.20 m, - 1.76 m)
6b. Find the motorcycle’s distance from the edge of the cliff position after 0.60 seconds.
A.
(1) Let r = distance
(2) r = [(7.20 m)2 + (- 1.76 m)2] 1/2
(3) r = 7.42 m
7. A batter hits a baseball so that it leaves the bat with an initial speed vo = 35.0 m/s at an initial angle
q = 47.9o.
7a. Find the position of the ball at 2.00 seconds
A.
(1) The ball is probably struck a meter or so above the ground, but we neglect this distance and assume that it starts at ground level.
(2) Vx = V cos q
(3) Vx = (35.0 m/s)(cos 47.9o)
(4) Vx = 23.46 m/s
(5) Voy = V sin q
(6) Voy = (35.0 m/s)(sin 47.9o)
(7) Voy = 25.97 m/s
(8) Vx = D x / D t
(9) D x = (Vx)(D t)
7a. (continued...)
(10) D x = (23.46 m/s)(2.00 s)
(11) D x = 46.9 m
(12) D y = Voyt - ½ gt2
(13) D y = (25.97 m/s)(2.00 s) – ½ (9.8 m/s2)(2.00 s)2 = 32.34 m
(14) D y = 32.34 m
(15) P(x,y) = (46.9 m, 32.3 m)
7b. Find the magnitude and direction of its velocity at 2.00 seconds
A.
(1) Vx = 23.46 m/s
(2) Vy = Voy – gt
(3) Vy = 25.97 m/s – (9.8 m/s2)(2.00 s)
(4) Vy = 6.37 m/s
(5) V = [(Vx)2 + (Vy)2] ½
(6) V=[(23.46 m/s)2 + (6.37 m/s)2] ½ = 24.31 m/s
(7) q = tan-1 (Vy / Vx)
(8) q = tan-1 (6.37 / 23.46) = 15.1o
(9) Velocity: 24.3 m/s, 15.1o north of x axis
7c. Find the time when the ball reaches its highest point.
A.
(1) Vy = Voy - gt
(2) 0 = Voy – gt
(3) Voy = gt
(4) t = Voy / g
(5) t = (25.97 m/s) / (9.8 m/s2)
(6) t = 2.65 s
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