physics - DOC by bjaylopez

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									39. Is it true that an object dropped from rest falls three times farther in the second second
after being released than it does in the first second?

Given: Vo= 0 g = 9.8 m/ s2 t1 = 1s t2 = 2s
Required: y2 = 3y1
Solution: at t1 ;    y1 = Vot + ½ gt2
                     y1 = (0)t + ½ (9.8m/s2)(1s)2
                     y1 = 0 + 4.9m
                     y1 = 4.9m
         at t2;      y2 = Vot + ½ gt2
                     y2 = (0)t + ½ (9.8m/s2)(2s)2
                     y2 = 0 + 19.6m
                     y2 = 19.600m
       therefore:    y2 = 3y1
                     19.600m = 3(4 .9m)
                     19.600m = 14.700m an object dropped from rest falls 3 times
                                            farther in the second second after being
                                            released than I t does in the first second!

40. How fast must a ball be thrown upward to reach a height of 12m?

Given: y = 12m Vf = 0 g = 9.8 m/s2
Required: Vo = velocity of the ball when released
Solution:             Vo2 = Vf2 – 2gy
                      Vo2 = 0 – 2 (- 9.8m/s2) (12m)
                      Vo = √235.2m/s
                      Vo = 15.336 m/s

41. Divers in Acapulco leap from a point 36m above the sea. What is their velocity when
they enter the water?

Given: y = 36m Vo = 0 g = 9.8m/s2
Required; Vf = the velocity when the diver enter the water
Solution:             Vf2 = Vo2 + 2gy
                          = (0)2 + 2(9.8m/s2)(36m)
                      Vf = √705.6m2/s2
                      Vf = 26.563m/s

42. A stone is dropped from a cliff 490m above its base. How long does the stone take to
fall?

Given: y = 490m         g = 9.8m/s2 Vo = 0
Required: t = time it takes for the stone to take to fall
Solution:                    y = Vot + ½ gt2
                        490m = (0)t + ½ (9.8m/s2)t2
                            t2 = (490m)/(4.9m/s2)
                        t = √100s2
                        t = 10s

43. A ball dropped from the roof of a building takes 4.0 s to reach the street. How many
feet high is the building?

Given: t = 4.0s      g = 9.8m/s2 Vo= 0
Required: y = how high in feet is the building
Solution:            y = Vot + ½ gt2
                       = (0)t + ½ (9.8m/ss)(4s)2
                       = ½ (156.8m)
Converting in ft;    y = 78.4m x (3.281ft)/(1m) = 257.230ft

44. A bullet is fired vertically upward and returns to the ground in 20s. Find the height it
reaches.

Given: t = 20s         g = 9.8m/s2
Required: y = total height that the bullet reached; Vo = initial velocity of the bullet
Solution:      Since we only need the total height then t2 = 10s and Vf = 0

                Vo = Vf – gt
                   = 0 – (-9.8m/ss)(10s)
                   = 98m/s

                Computing for y;

                y = Vot + ½ gt2
                  = (98m/s)(10s) + ½ (-9.8m/s2)(10s)2
                  = 490m

45. Find the initial and final velocities in ft/s of a ball thrown vertically upward that that
returns to the thrower 3.0s later.

Given: t = 3s           g = 9.8m/s2
Required: Vo = initial velocity of the ball in ft/s
           Vf = final velocity of the ball in return
Solution:      Converting g = 9.8m/s x (3.281ft)/(1m) = 32.154ft/s2
               Calculating ball traveling upward then t1 = ½ t or t1= 1.5s and Vf1= 0
               Vf1 = Vo + gt1
                 0 = Vo + (-32.154ft/s2)(1.5s)
               Vo = 48.231ft/s
               Then Vf = 48.231ft/s because Vo = Vf
46. A stone is thrown vertically upward at 9.8 m/s. when it will reach the ground?

Given: Vo = 9.8m/s g = 9.8m/s2
Required: t = total time traveled by the stone

Solution: Calculating half of the traveled time t1
              Vf = Vo + gt
              0 = (9.8m/s) + (-9.8m/s2)t1
              t1 = 1s
              then total time t = 2t1; t = 2(1s) t = 2s

47. A ball is thrown vertically downward at 10m/s. What is it velocity 1.0s later? 2.0s
later?

Given: t1 = 1s         t2 = 2s       g = 9.8m/s2          Vo = 10m/s
Required: Vf1 = velocity at t1
           Vf2 = velocity at t2
Solution:      Vf1 = Vo + gt1
                   = (10m/s) + (9.8m/s2)(1s)
               Vf1 = 19.800m/s

               Vf2 = Vo + gt2
                   = (10m/s) + (9.8m/s2)(2s)
               Vf2 = 29.6m/s

48. A ball is thrown vertically upward at 10m/s. What is its velocity 1.0s later? 2.0s later?

Given: t1 = 1s         t2 = 2s       g = 9.8m/s2          Vo= 10m/s
Required: Vf1 = velocity at t1
           Vf2 = velocity at t2
Solution:      Vf1 = Vo + gt1
                   = (10m/s) + (-9.8m/s2)(1s)
                   = 0.200m/s
               Vf2 = Vo + gt2
                   = (10m/s) + (-9.8m/s2)(2s)
                   = -9.6m/s

49. A lead pellet is propelled vertically upward by an air rifle with an initial velocity of
16m /s. Find the maximum height.
Given:         g = 9.8m/s2 Vo= 16m/s
Required: y = maximum height of the lead pellet
Solution:               Vf2 = Vo2 + 2gy
                         0 = (16)2 + 2(-9.8m/s2)y
                         y = (256m2/s2)/(19.6m/s2)
                         y = 13.061m
50. At what velocity should a stone be thrown downward from a cliff 120m high to reach
the ground in 4.0s?

Given:        t = 4s y = 120m        g = 9.8m/s2
Required: Vo = velocity of the stone
Solution:                   y = Vot + ½ gt2
                      120m = Vo(4s) + ½ (9.8m/ss)(4s)2
                       (4s)Vo =120m-78.4m
                       (4s)Vo = 41.6m
                           Vo = 10.400m/s

51. An apple is thrown vertically downward from a cliff 48m high and reaches the ground
2.0s later. What was the apple’s initial velocity?

Given:        t = 2s y = 48m           g = 9.8m/s2
Required: Vo = apple’s initial velocity
Solution:                  y = Vot + ½ gt2
                      48m = Vo(2s) + ½ (9.8m/ss)(2s)2
                         Vo = (48m – 19.6m)/(2s)
                         Vo = 14.200m/s

52. From Fig. 1.16 we can see that the air resistance experienced by a falling object is not
an important factor until a speed of about half its terminal speed is reached. The terminal
speed of a golf ball is 40m/s. How much time is needed for a dropped goofball to reach to
reach a speed of half this? How far does it fall in this time?

Given: Vt = 40m/s       Vt = 2Vf then Vf = 20m/s
Required:

53. The acceleration of gravity at the surface of Mars is 3.7m/s2. A stone thrown upward
on Mars reaches a height of 15 m. (a) Find its initial velocity. (b) What is the total time of
flight?

Given:        y = 15m          g = 3.7m/s2
Required: Vo = initial velocity of the stone; t1= total time of the stone’s flight
Solution:              Vf2 = Vo2 + 2gy
                        0 = Vo 2 + 2(-3.7m/s2)(15m)
                        Vo = √111m2/s2
                        Vo = 10.536 m/s

                     Vf1 = Vo + gt1
                         0 = (10.536m/s) + (-3.7m/s2)t1
                        t2 = 2.848s
               computing for the total time
                     t1 = 2t1 then t1 = 2(2.848s) = 5.695s
54. A movie scene in which a car is supposed to fall from a cliff 36m high is filmed using
a model car that falls from a model cliff 1.0m high. The film shown at 24 frames/s. How
fast should the camera be run for the result to be realistic?

55. In a novel a pirate forces captured a sailor to “walk the plank” (something unknown
in the history of piracy), from which he falls into the sea 7.0m below. The author claims
the sailor’s entire life flashes before him during his fall. (a) How much time is available
for this? (b) With what velocity will the sailor enter the water?

Given: y = 7m         g = 9.8m/s2
Required: t1 = time available for the sailor Vf = velocity of the sailor when he enters the
water
Solution:                  y = Vot + ½ gt2
                        7m = 0t + ½ (9.8m/ss)t2
                          t = √1.429
                          t = 1.195s

                          Vf = Vo + gt
                          Vf = 0 + ( 9.8m/s2)(1.195)2
                          Vf = 11.71m/s

56. A bagel is thrown vertically upward with a velocity of 60 ft/s from a bridge. On the
way down the bagel just misses the bridge and falls into the water 5.0s after having been
thrown. Find the height of the bridge above the water?

Given: Vo = 60ft/s x (1m)/(3.281ft) = 18.287m/s     t= 5s
Required: y1 = height of the bridge over the water
Solution: t1 = time when the bagel reach the height after is has been thrown
                Vf1 = Vo + gt1
                 0 = (18.287m/s) + (9.8m/s2)t
                 t1 = 1.866s

                t2 = total time when the bagel returns to the height of the bridge
                t2 = 2t1 then t2 = 2(1.866s) = 3.732s

                t3 = time that takes when the bagel falls from the height of the bridge to
                the water
                t3 = t – t2 then, t3 = 5s – 3.732s so, t3 = 1.268s

                           y = Vot + ½ gt2
                             = ( 18.287m/s)(1.268s) + ½ (9.8m/ss)(1.268s)2
                             = 23.188m + 7.878m
                             = 31.066m
57. A British parachutist bails out at an altitude of 150m and accidentally drops his
monocle. If he descends at the constant velocity of 6.0m/s, how much time separates the
arrival of the monocle on the ground from the arrival of the parachutist himself?

58. A Russian balloonist floating at an altitude of 150m accidentally drop his samovar
and starts to ascend at the constant velocity of 1.2m/s. How high will the balloon be when
the samovar reaches the ground?

Given: y = 150m                Vo = 1.2m/s
Required: y2 = height of the balloon when the samovar reaches the ground
Solution:          y = Vot + ½ gt2
                       150m = (0)t + ½ (9.8m/ss)t2
                             t = 5.533s; time when the samovar reaches the ground
               y3 = height ascended by the balloon

                       y3 = ( Vo – Vf ) t
                                 2
                       y3 = {(1.2m/s)/2}(5.533)
                       y3 = 3.3198m
               y2 = y – y3
                  = 150m-3.3198m
                  = 146.680m

59. An orangutan throws a coconut vertically upward at the foot of a cliff 40m high while
his mate simultaneously drops another coconut from the top of the cliff. The two
coconuts collide at an altitude of 20m. What was the initial velocity of the coconut that
was thrown upward?

60. When a flea jumps, it accelerates through about 0.80mm (a little less than the legs)
and is able to reach a height of as much as 10cm. (a) Find the flea’s acceleration
(assumed constant) and its velocity at takeoff when this occurs. (b) When a person jumps,
the acceleration distance is about 50cm. If the acceleration of a person were the same as
that of a flea, find the velocity at takeoff and the height that would be reached.

Given:         t = 4s         g = 9.8m/s2


61. A helicopter is climbing at 8.0m/s when it drops a pump near a leaking boat. The
pump reaches the water 4.0s afterward. How high was the helicopter when the pump was
dropped? When the pump reached the water?

Given:         t = 4s Vo = -8m/s      g = 9.8m/s2
Required: y = height of the helicopter when the pump was dropped
Solution:             y = Vot + ½ gt2
                         = (-8m/s)(4s) + ½ (9.8m/ss)(4s)2
                      y = 46.400m
62. Figure 1.25 is a plot of displacement versus time for a beetle moving along a straight
stick. (a) Plot the beetle’s velocity versus time. (b) What is the beetle’s average speed
during the same period?

63. A Snake is slithering toward you at 1.5m/s. If you start walking when it is 5.0 m
away, how fast must you go in order that snake not overtake you when you have gone
40.0m?

64. A car covers one-quarter of the distance to the next town at 40km/h; another quarter
at 50km/h and the rest 90 km/h. find the car’s average speed for the entire distance.

65. A man drives one-third of the distance home at 25mi/h. How fast must he drive the
rest of the way in order to average 35mi/h for the entire trip?

66. A woman on a bicycle a hill at 4.0m/s and then returns down the same hill at 8.0m/s.
What is her average speed for the entire trip?

67. The odometer of a car is checked at 1.0 min intervals, and the readings below are
obtained. Calculate the speed of the car in each time interval and plot the results on a
graph. Describe the motion of the car with the help of this graph.

Time (min)             0               1.0             2.0             3.0
Distance(mi)           42.20           42.74           43.64           44.90

Time(min)              4.0             5.0             6.0             7.0
Distance(mi)           46.34           47.78           49.22           50.66

Time(min)              8.0             9.0             10.0
Distance(mi)           51.74           52.10           52.10

68. A European train passes successive kilometer posts at the times given below. Plot the
data on a graph and determine whether the train’s speed is constant over the entire
distance. If it is not constant, plot the train’s speed in each time interval versus time and
find the acceleration. What are the train’s initial and final speeds?

69. A car that is leaking oil at a constant rate is uniformly accelerated from rest starting
out just when a drop falls. The first drop down the road is 0.80m from the car’s starting
point. How far from the starting point is the fourth drop?

70. A sprinter accelerates from rest until he reaches a velocity of 40ft/s and then
continues running with this velocity. If he takes 11.0s to cover 300ft what was his
acceleration and how long did it last?
71. A bus travels 400m between two stops. It starts from rest and accelerates at 1.5 m/s2
until it reaches a velocity of 9.0m/s. The bus continues at this velocity and then
decelerates at 2.0m/s2 until comes to a halt. Find the total time required for the journey.

72. A rocket is launched upward with an acceleration of 100m/s2. Eight seconds later the
fuel exhausted. Find (a) the highest velocity the rocket attains; (b) the maximum altitude;
(c) the total time of flight; and (d) the velocity with which the rocket strikes the ground.

73. A girl throws a ball vertically upward at 10m/s from the roof of a building 20m high.
(a) How long will it take the ball to reach the ground? (b) What will its velocity be when
it strikes the ground?

74. A girl throws a ball vertically downward at 10m/s from the roof of a building 20m
high. (a) How long will it take the ball to reach the ground? (b) What will its velocity be
when it strikes the ground?

75. An elevator has a maximum acceleration of + 1.5 m/s2 and a maximum velocity of
6.0m/s. Find the shortest period of time required for it to take a passenger to the tenth
floor of a building from street level, a height of 50m, with the elevator coming to a stop at
this floor.

76. A person in an elevator drops an apple from a height 2.0m above the elevator’s floor
and , with a stopwatch, time the fall of the apple to the floor. What is found (a) when the
elevator is ascending with an acceleration of 1.0m/s2; (b) when it is descending with an
acceleration of 1.0m/s2; (c) when it is ascending at a constant velocity of 3.0m/s; (d)
when it is descending at a constant velocity of 3.0m/s; and (e) when the cable has broken
and it is descending at free fall?

77. When a certain ball is dropped to the ground from a height of 2.0m, it bounces back
up to its original height. (a) Plot a graph of the height of the ball versus time for the first
3.0s of its motion. Plot a graph of the ball’s velocity during the same period.

78. A champagne bottle is held upright 4ft above the floor as the wire around its cork is
remove. The cork then pops out, rises vertically, and falls to the floor 1.4s later. (a) What
height above the bottle did the cork reach? (b) What was the cork’s initial velocity? (c) Its
final velocity?

								
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