Homework



Homework

1. Which one of the following quantities is NOT a vector?

|a. 10 mi/hr, east |b. 10 mi/hr/sec, west |c. 35 m/s, north |d. 20 m/s |

2. Two vector are added together, and their vector sum is zero, what can you say about the magnitude and direction of the two initial vectors?

3. Two vectors have magnitudes d1 = 3.5 km and d2 = 4.0 km. What are the maximum and minimum magnitudes of their vector sums?

4. A delivery truck travels 14 m north, 16 m east and 26 m south. What is its final displacement from the origin (magnitude and direction)?

5. A car is driving 125 km west and then 65 km 45o south of west. What is the displacement of the car form the point of origin (magnitude and direction)? Draw a diagram.

6. Three vectors are shown in the diagram below. Determine the sum of the three vectors. Their magnitudes are given in arbitrary units. Give the resultant in terms of (a) components (b) magnitude and angle with x axis.

(So not be scared by the format of the question or the presence of a third vector, just simply break the vectors into their components (x and y) and add or subtract the components on the same plane depending on their direction, then using the resultant x and y summed totals find the overall resultant)

7. A man in a rowboat is trying to cross a river that flows due west with a strong current. The man starts on the south bank and is trying to reach the north bank directly north from his starting point. He should:

a. head due north

b. head due west

c. head in a northwesterly direction

d. head in a northeasterly direction

8. Huck Finn walks at a speed of 1.0 m/s across his raft (that is, he walks perpendicular to the raft’s motion relative to the shore). The raft is traveling down the Mississippi River at a speed of 2.7 m/s relative to the river back. What is the velocity (speed and direction) of Huck relative to the river bank? (**Be sure to draw and label a diagram.)

9. A boat’s speed in still water is 1.85 m/s and if it heads directly across the stream whose current is 1.20 m/s (a) What is the velocity (magnitude and direction) of the boat relative to the shore? (b) If the river is 110 m wide, how long will it take to cross and how far downstream will the boat be then?

10. A motorboat whose speed in still water is 3.60 m/s must aim upstream at an angle of 27.5o(with respect to a line perpendicular to the shore) in order to travel directly across the stream. (a) What is the speed of the current? (b) What is the resultant speed of the boat with respect to the shore?

11. A boat whose speed in still water is 2.20 m/s must cross a 260-m-wide river and arrive at a point 110 m upstream from where it starts. To do so, the pilot must head the boat at a 45o angle upstream. What is the speed of the river’s current?

D2

12. An airplane, whose air speed is 600 km/h, is supposed to fly in a straight path 35o north of east. But a steady 100 km/h wind is blowing from the north. In what direction should the plane head? (Hint: resolve the initial air speed into its vector components and then using a diagram, determine how the plane would compensate for the wind speed. Add this to the appropriate vector component and using trigonometry determine the new angle of the new resultant)

13. Two rowers, who can row at the same speed in still water, set off across the river at the same time. One heads straight across and is pulled downstream somewhat by the current. The other one heads upstream at an angle so as to arrive at a point opposite the starting point. Which rower reaches the opposite side first? Why? (KNOW HOW TO ANSWER QUESTIONS LIKE THIS-VERY SIMILAR TO SOME PROVINCIAL QUESTIONS)

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14. Use the principle of slope to describe the motion of the objects depicted by the two plots below. In your description, be sure to include such information as the direction of the velocity vector (i.e., positive or negative), whether there is a constant velocity or an acceleration, and whether the object is moving slow, fast, from slow to fast or from fast to slow. Be complete in your description.

[pic]

15. Test your understanding of slope calculations by determining the slope of the line below.

[pic]

16. Determine the velocity (i.e., slope) of the object as portrayed by the graph below.

[pic]

17. Consider the graph at the right. The object whose motion is represented by this graph is ... (include all that are true):

1. moving in the positive direction.

2. moving with a constant velocity.

3. moving with a negative velocity.

4. slowing down.

5. changing directions.

6. speeding up.

7. moving with a positive acceleration.

8. moving with a constant acceleration.

18. The velocity-time graph for a two-stage rocket is shown below. Use the graph and your understanding of slope calculations to determine the acceleration of the rocket during the listed time intervals.

1. t = 0 - 1 second

2. t = 1 - 4 second

3. t = 4 - 12 second

[pic]

19. Describe the motion depicted by the following velocity-time graphs. In your descriptions, make reference to the direction of motion (+ or - direction), the velocity and acceleration and any changes in speed (speeding up or slowing down) during the various time intervals (e.g., intervals A, B, and C).

[pic]

[pic]

[pic]

20. Now try the following two practice problems as a check of your understanding.

[pic]

 

21. Now try the following two practice problems as a check of your understanding.

[pic]

22. Now try the following two practice problems as a check of your understanding.

[pic]

23.  Draw two graphs, one a position vs time graph and one a velocity verses time graph for a free falling object.

24. An airplane accelerates down a run-way at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before take-off.

25. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

26. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.

27. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.

28. An engineer is designing the run-way for an airport. Of the planes which will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The lowest take-off speed will be 65 m/s. Assuming these minimum parameters for the same plane, what is the minimum allowed length for the run-way?

29. A kangaroo is capable of jumping to a height of 2.62 m. Determine the take-off speed of the kangaroo.

30. If Michael Jordan has a vertical leap of 1.29 m, then what is his take-off speed and his hang time (total time to move upwards to the peak and then return to the ground)?

31. A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)

32. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.

33. It was once recorded that a Jaguar left skid marks which were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.

34. Otto Emissions is driving his car at 25.0 m/s. Otto accelerates at 2.0 m/s2 for 5 seconds. Otto then maintains a constant velocity for 10.0 more seconds.

1. Represent the 15 seconds of Otto Emission's motion by sketching a velocity-time graph. Use the graph to determine the distance which Otto traveled during the entire 15 seconds.

2. Finally, break the motion into its two segments and use kinematic equations to calculate the total distance traveled during the entire 15 seconds.

35. Luke Autbeloe, a human cannonball artist, is shot off the edge of a cliff with an initial upward velocity of +40.0 m/s. Luke accelerates with a constant downward acceleration of -10.0 m/s2 (an approximate value of the acceleration of gravity).

1. Sketch a velocity-time graph for the first 10 seconds of Luke's motion.

2. Use kinematic equations to determine the time required for Luke Autbeloe to drop back to the original height of the cliff. Indicate this time on the graph.

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36. Consider these diagrams in answering the following questions.

[pic]

Which diagram (if any) might represent ...

1. The initial horizontal velocity?

2. The initial vertical velocity?

3. The horizontal acceleration?

4. The vertical acceleration?

5. The net force?

37. Suppose an airplane drops a flare while it is moving with a constant horizontal speed at an elevated height. Assuming that air resistance is negligible, where will the flare land relative to the plane?

1. directly below the plane

2. below the plane and ahead of it

3. below the plane and behing it.

38. A cannonball is launched horizontally from the top of an 80-meter high cliff. How much time will it take for the ball to reach the ground and at what height will the ball be after each second of travel? (Assume 10 m/s2 = acceleration of gravity to simplify calculations)

39. The diagram below shows the trajectory for a projectile launched non-horizontally from an elevated position on top of a cliff. The initial horizontal and vertical components of the velocity are 8 m/s and 20 m/s as shown in the diagram. Positions of the object at 1-second intervals are shown. determine the horizontal and vertical velocities at each instant shown in the diagram. (Assume 10 m/s2 = acceleration of gravity to simplify calculations)

[pic]

40. Utilize kinematic equations and projectile motion concepts to fill in the blanks in the following tables.

[pic]

41. A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

42. A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

43. Omiachen Head, an archer, is riding in the back of a pickup truck which is traveling at a constant velocity of 40 miles per hour. He decides to fire off his last arrow by shooting it straight up into the air. He reasons to himself that if he shoots the arrow straight up, it will come down at the same location from which it was shot. This way he would be far away from the arrow when it came down. After all, he's moving horizontally at 40 miles an hour. He'll be far away. Right? Explain.

44. A cat is dropped (accidentally of course) from the top of a building that is 25 meters high.

a. What is the initial vertical velocity of the cat?

b. Determine the time it takes the poor cat to reach the pavem...OK safety net 25 meters below.

c. Determine the velocity of the cat as she strikes the safety net.

The cat survives, scratches her owner and is throw horizontally with a velocity of 15 m/s (not so accidentally) from the top of a building that is 25 meters high.

d. What is the initial vertical velocity of the cat?

e. Determine the time it takes the poor cat to reach the pavem...OK safety net 25 meters below.

f. Determine how far the cat will travel horizontally by the time it has fallen 25 meters (again to place the safety net).

g. What will be the cat’s final impact velocity (magnitude and direction)?

45. Divers at Acapulco dive from a cliff that is 61 meters high. If the rocks below the cliff extend outward for 23 meters, what is the minimum horizontal velocity a diver must have to clear the rocks?

46. Harry the human cannonball is now fired with a velocity of 30.0 m/s at an angle of 60.0o with the horizontal. Determine Harry's:

a. initial vertical and horizontal velocity components.

b. total time in the air.

c. horizontal distance traveled (where the net is hopefully located).

d. maximum height achieved (which is hopefully lower than the ceiling).

47. A football is kicked with a velocity of 38.0 m/s at an angle of 40.0 degrees. Assume the ball takes off and lands at the same height.

a. What is the direction and magnitude of the ball velocity vector after 0.20 seconds?

b. Find the horizontal range the ball travels.

48. A golf ball lands on an elevated green with a final height of 14.0 m, a horizontal velocity of 28.0 m/s and a vertical velocity of -18.0 m/s. The height of the ball at takeoff was 3.00 m. How far did the golf ball travel?

49. A javelin is thrown with an initial height of 1.6 m and a vertical velocity of 27 m/s and a horizontal velocity of 30 m/s. How high is the javelin off the ground after 2.0 seconds? What is the horizontal and vertical velocity of the javelin after 1.5 seconds?

50. A water balloon is launched with a velocity of 46.0 m/s at an angle of 40.0° with an initial height of .800 m. The balloon smashes into a building which is a horizontal distance of 15.0 meters from where the balloon was launched. What is the horizontal and vertical velocity of the balloon when it hits the building? How high is the balloon when it hits the building? How many seconds does it take for the balloon to hit the building? What is the direction of the balloons velocity vector when it hits the building?

51. A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile strikes the top of the cliff 7.6 seconds after it is fired, find the initial velocity of the projectile (magnitude and direction).

Homework Answers

1: d A vector has both magnitude and direction. Since d has only magnitude it is a scalar quantity.

2: They must have been equal in magnitude but opposing opposites in direction.

3: Their maximum is their sum if they were placed in a straight line head to tail = 7.5 km Their minimum is their difference if they were placed in a straight line but with one in one direction and the other in the opposite direction = -0.5 or .5 depending on the order of displacement from the origin.

4: 20 m, 53o E of S or 37oS of E (be aware that these angles can be written in more than one way, draw a diagram if you are unsure if you answer is right if it varies from the ones listed below)

5: 177 km, 15o S of W

6: (a) 35.9, 17.3 (b) 39.9, 25.8o above the x-axis

7: The current will drag the boat westward, so to counteract that motion the boat must head in a northeasterly direction. The actual angle depends on the strength of the current and how fast the boat moves relative to the water. If the current is weak and the rower is strong, then the boat can head almost, but not quite, due north.

8: 2.9 m/s, 20o from the river bank.

9: (a) 2.21m/s, 57o from the river bank (b)60 s and 72 m

10: (a)1.66 m/s (b) 3.19 m/s

11: 0.896 m/s

12: 42.9o N of E

D3

13:Rower 1 reaches the opposite side first. Why? Because Rower 2’s up stream velocity is resolved into two components, one against the river and the other across the river. This means the across river velocity will be smaller than rower 1’s. Since rower 2’s initial across velocity is reduced rower two must take longer to cross the river.

14: (A) positive or rightward velocity (+ slope)

changing velocity = accelerating (changing slope)

moving slow to fast (slope gradual to steep)

(B) negative or leftward velocity (- slope)

changing velocity = acceleration (changing slope)

moving slow to fast (slope gradual to steep)

15: Using the two data points, the rise =-24.0 m (df – di = Δd = 2.0 – 26.0 = -24.0 m) and the run =8.0 s (tf – ti = Δt = 8.0 – 0 = 8.0 s), thus the slope = -3 m/s

16: The velocity (slope) = 4 m/s.

If you think the slope is 5m/s then you’re making a common mistake. You are picking one point (probably 5s,25m) and dividing y/x instead you must pick two points and divide the change in y by change in x.

17: a,d, h apply

18: (a) The acceleration = +40 m/s2 or +40 m/s/s (Remember to use a slope calculation)

(b) the acceleration = +20 m/s2

(c) the acceleration = -20 m/s2

19: (A) the object moves in a + dir’n at a constant speed – zero acceleration (Interval A). The object then continues in the + dir’n while slowing down with a neg. accel’n (Interval B). Finally, the object moves at a constant speed in the + dir’n, slower than before (Interval C).

(B) The object moves in the + dir’n while slowing down – a neg. accel’n (Interval A). It then remains at rest (Interval B). The object then moves in the – dir’n while speeding up – a neg. accel’n (Interval C)

(C) The object moves in the + dir’n with a constant velocity and zero accel’n (Interval A). The object then slows down while moving in the + dir’n – a neg. accel’n until it finally reaches a zero velocity (stops) (Interval B). Then the object moves in the – dir’n while speeding up – a neg accel’n (Interval C).

20: (A) The object was displaced 120 m during the first 4 seconds

(B) The object was displaced 90 m during the time 3 to 6 seconds

21: (A) The object was displaced 5 m during the first second

(B) The object was displaced 45 m during the first 3 seconds

22: (A) The object was displaced 25 m during the time 2 to 3 seconds

(B) The object was displaced 40 m during the first 2 seconds

23:

[pic][pic]

24:

|Given: |Find: |

|a = +3.2 m/s2 |d = ?? |

|t = 32.8 s | |

|vi = 0 m/s | |

| | |

1. d = vi*t + 0.5*a*t2

2. d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2

3. d = 1720 m

25:

|Given: |Find: |

|a = -9.8 m |d = ?? |

|t = 2.6 s |vf = ?? |

|vi = 0 m/s | |

| | |

1. d = vi*t + 0.5*a*t2

2. d = (0 m/s)*(2.6 s)+ 0.5*(-9.8 m/s2)*(2.6 s)2

3. d = 33 m

4. vf = vi + a*t

5. vf = 0 + (-9.8 m/s2)*(2.6 s)

6. vf = -25.5 m/s (- indicates direction)

26:

|Given: |Find: |

|vi = 0 m/s |t = ?? |

|d = -1.40 m | |

|a = -1.67 m/s2 | |

| | |

1. d = vi*t + 0.5*a*t2

2. -1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2

3. -1.40 m = 0+ (-0.835 m/s2)*(t)2

4. (-1.40 m)/(-0.835 m/s2) = t2

5. 1.68 s2 = t2

6. t = 1.29 s

27:

|Given: |Find: |

|vi = 0 m/s |a = ?? |

|vf = 7.10 m/s | |

|d = 35.4 m | |

| | |

1. vf2 = vi2 + 2*a*d

2. (7.10 m/s)2 = (0 m/s)2 + 2*(a)*(35.4 m)

3. 50.4 m2/s2 = (0 m/s)2 + (70.8 m)*a

4. (50.4 m2/s2)/(70.8 m) = a

5. a = 0.712 m/s2

28:

|Given: |Find: |

|vi = 0 m/s |d = ?? |

|vf = 65 m/s | |

|a = 3 m/s2 | |

| | |

1. vf2 = vi2 + 2*a*d

2. (65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

3. 4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d

4. (4225 m2/s2)/(6 m/s2) = d

5. d = 704 m

29:

|Given: |Find: |

|a = -9.8 m/s2 |vi = ?? |

|vf = 0 m/s | |

|d = 2.62 m | |

| | |

1. vf2 = vi2 + 2*a*d

2. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)

3. 0 m2/s2 = vi2 - 51.35 m2/s2

4. 51.35 m2/s2 = vi2

5. vi = 7.17 m/s

30:

|Given: |Find: |

|a = -9.8 m/s2 |vi = ?? |

|vf = 0 m/s |t = ?? |

|d = 1.29 m | |

| | |

1. vf2 = vi2 + 2*a*d

2. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(1.29 m)

3. 0 m2/s2 = vi2 - 25.28 m2/s2

4. 25.28 m2/s2 = vi2

5. vi = 5.03 m/s

6. To find hang time, find the time to the peak and then double it.

7. vf = vi + a*t

8. 0 m/s = 5.03 m/s + (-9.8 m/s2)*tup

9. -5.03 m/s = (-9.8 m/s2)*tup

10. (-5.03 m/s)/(-9.8 m/s2) = tup

11. tup = 0.513 s

12. hang time = 1.03 s

31:

|Given: |Find: |

|vi = 367 m/s |a = ?? |

|vf = 0 m/s | |

|d = 0.0621 m | |

| | |

1. vf2 = vi2 + 2*a*d

2. (0 m/s)2 = (367 m/s)2 + 2*(a)*(0.0621 m)

3. 0 m2/s2 = (134689 m2/s2) + (0.1242 m)*a

4. -134689 m2/s2 = (0.1242 m)*a

5. (-134689 m2/s2)/(0.1242 m) = a

6. a = -1.08*106 m /s2

7. (The - sign indicates that the bullet slowed down.)

32:

|Given: |Find: |

|a = -9.8 m/s2 |d = ?? |

|t = 3.41 s | |

|vi = 0 m/s | |

| | |

1. d = vi*t + 0.5*a*t2

2. d = (0 m/s)*(2.6 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2

3. d = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2)

4. d = -57.0 m

5. (NOTE: the - sign indicates direction)

33:

|Given: |Find: |

|a = -3.90 m/s2 |vi = ?? |

|vf = 0 m/s | |

|d = 290 m | |

| | |

1. vf2 = vi2 + 2*a*d

2. (0 m/s)2 = vi2 + 2*(-3.90 m/s2)*(290 m)

3. 0 m2/s2 = vi2 - 2262 m2/s2

4. 2262 m2/s2 = vi2

5. vi = 47.6 m /s

34:

a. The velocity-time graph for the motion is:

[pic]

The distance traveled can be found by a calculation of the area between the line on the graph and the time axis. This area would be the area of the triangle plus the area of rectangle 1 plus the area of rectangle 2.

Area = 0.5*b1*ht + b1*h1 + b2*h2

Area = 0.5*(5.0 s)*(10.0 m/s) + (5.0 s)*(25.0 m/s) + (10.0 s)*(35.0 m/s)

Area = 25 m + 125 m + 350 m

Area = 500 m

b. The distance traveled can be calculated using a kinematic equation. The solution is shown here.

First find the d for the first 5 seconds:

|Given: |Find: |

|vi = 25.0 m/s |d = ?? |

|t = 5.0 s | |

|a = 2.0 m/s2 | |

| | |

d = vi*t + 0.5*a*t2

d = (25.0 m/s)*(5.0 s) + 0.5*(2.0 m/s2)*(5.0 s)2

d = 125 m + 25.0 m

d = 150 m

Now find the d for the last 10 seconds:

|Given: |Find: |

|vi = 35.0 m/s |d = ?? |

|t = 10.0 s | |

|a = 0.0 m/s2 | |

| | |

(Note: the velocity at the 5 second mark can be found from knowing that the car accelerates from 25.0 m/s at +2.0 m/s2 for 5 seconds. This results in a velocity change of a*t = 10 m/s, and thus a velocity of 35.0 m/s.)

d = vi*t + 0.5*a*t2

d = (35.0 m/s)*(10.0 s) + 0.5*(0.0 m/s2)*(10.0 s)2

d = 350 m + 0 m

d =350 m

The total distance for the 15 seconds of motion is the sum of these two distance calculations (150 m + 350 m): distance = 500 m

35:

1. The velocity-time graph for the motion is:

[pic]

2. The time to rise up and fall back down to the original height is twice the time to rise up to the peak. So the solution involves finding the time to rise up to the peak and then doubling it.

|Given: |Find: |

|vi = 40.0 m/s |tup = ?? |

|vf = 0.0 m/s |2*tup = ?? |

|a = -10.0 m/s2 | |

| | |

vf = vi + a*tup

0 m/s = 40 m/s + (-10 m/s2)*tup

(10 m/s2)*tup = 40 m/s

tup = (40 m/s)/(10 m/s2)

tup = 4.0 s

2*tup = 8.0 s

36: (a) the initial horizontal velocity is A (it’s the only horiz. Velocity)

(b) the initial velocity could be B (if projected down) or C (if projected upward)

(c) none of these, there is no horizontal acceleration

(d) the vertical acceleration is D; it is always downwards

(e) the net force on a projectile is down (there is only one force – gravity, and its downwards)

37: (a) the horizontal motion of the falling flare remains constant, and as such, the flare will always be positioned directly below the plane. The force of gravity causes the flare to fall but does not effect its horizontal motion.

38: It will take 4 seconds to fall 80 meters. (y=0.5*g*t2) g= 10 m/s2 (for diagram purposes) y = 80m.

At t = 1s , y = 5 m so height is 75 m (80m - 5m)

At t = 2s , y = 20 m so height is 60 m (80m - 20m)

At t = 3s , y = 45 m so height is 35 m (80m - 45m)

At t = 4s , y = 80 m so height is 0 m (80m - 80m)

Note: The cannonball’s horizontal speed does not effect the time to fall a vertical distance of 80 m.

39: The vx values will remain 8 m/s for the entire 6 seconds

The vy values will be changing by 10 m/s each second (approximated for diagram purposes) thus vy = 10 m/s (t = 1s), vy = 0 m/s (t = 2s), vy = -10 m/s (t = 3s), vy = -20 m/s (t = 4s), vy = -30 m/s (t = 5s), vy = -40 m/s (t = 6s),

40: A: 14.6 m B: 161 m C: 2.87 s D: 5.74 s E:41.2 m F: 235 m G: 19.2 m/s H: 16.1 m/s I: 1.61 s J: 3.21 s K: 79.9 deg L: 19.7 m/s M: 1.97 s N:3.94 s O:18.7 m/s P: 3.47 m/s Q: 3.47 s R: 6.95 s

41: Horiz info: x = 35.0 m

Vertical info: y = -22.0 m; v1y = 0 m/s; a = -9.8 m/s

Use y = viyt + 0.5at2 to solve for time. Time = 2.1189 s or 2.12 s rounded.

Now use x = v*t to solve for vx;

By subbing 35.0 m for x and 2.1189 s for t, the v can be found to be 16.5 m/s.

42: Initial hoizonatl componenets

vx = 10.6 m/s

vy = 5.6 m/s

horizontal info: x = ?; vx = 10.6

vertical info: y=?; viy = 5.6 m/s; vfy = -5.6 m/s; a = -9.8 m/s2

Use viy = vfy + at; time of flight is 1.1 or 1.1428 unrounded

Now use x = v*t; (time of flight = 1.1428, v = 10.6 m/s): x= 12.113 or rounded to 12.1 m

Finally use vfy2 = viy2 + 2a*y; (vfy = 0 at peak of height, viy = 5.6 m/s); y = 1.6 m

43: Since the arrow is shot from the back of a truck which is traveling at a velocity of 40 mi/hr, the arrow also has a horizontal velocity of 40 mi/hr. Once launched, the arrow will also have a vertical component of its velocity, but its horizontal component will remain at 40 mi/hr (unless air resistance comes into play). As a result, the horizontal displacement of the arrow will be the same as the truck since they are traveling at the same speed, in the same direction for the same amount of time...BONK!

44.

(a) Vyi=0m/s

(b)

[pic]

(c)

[pic]

(d) If thrown horizontally, the initital vertical velocity is 0 m/s.

(e)

[pic]

(f)

[pic]

(g)

Since we already know the cat’s horizontal velocity just before it hits the ground (i.e. same through out its fall) we can determine the cat’s vertical velocity before it hits the ground:

vyf2 = vyi2 + 2ady = 0 + 2(-9.8)(25) = -22.1359

Using the cat’s horizontal velocity (same as initial horizontal velocity) and the final vertical velocity of the cat after falling 25 meters, we can determine the final impact velocity using Pythagoras:

vf2 = vfy2 + vfx2 = (-22.1359)2 + (15)2 vf = 26.7395 – rounded to - 27 m/s

Remember, it’s a downward negative velocity so don’t forget to make your final answer negative.

45.

[pic]

46.

a. initial vertical and horizontal velocity components.

[pic]

b. total time in the air.

[pic]

c. horizontal distance traveled (where the net is hopefully located).

[pic]

d. maximum height achieved (which is hopefully lower than the ceiling).

[pic]

47. a. [v = 36.8 m/s, θ = 37.7]

b. [Δdx = 145 m]

48. [Δdx = 117 m]

49. [Δdy = 36 m, Vyf = 12 m/s, Vx = 30m/s]

50. [Vy = 25.4 m/s, Vx = 35.2 m/s, Δdy = 12.5 m, t = 0.426 s, θ = 35.8]

52. [63 m/s,66o above the horizontal]

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