Motion in One Dimension Problem F

Transcription

1 NAME DATE CLASS Motion in One Dimension Problem F FALLING OBJECT PROBLEM When it is completed in 2002, the International Financial Center in Taipei, Taiwan, will be the tallest building in the world. Suppose a construction worker on the top-most floor of the building accidentally knocks a wrench off a ledge. The wrench hits the ground below 9.56 s later. What is the distance between the top-floor of the International Financial Center and the ground. Assume there is no air resistance. SOLUTION. DEFINE Given: t = 9.56 s Unknown:? 2. PLAN 3. CALCULATE Choose the equation(s or situation: Displacement is unknown, as is the final velocity. Because time, acceleration, and initial velocity are known, the equation for displacement with constant acceleration can be used. v i t + 2 a t 2 Substitute the values into the equation(s and solve: (0 s(9.56 S + 2 ( 9.8 s 2 (9.56 s 2 (0 m + ( 448 m 448 m distance from top of building to ground = 448 m Copyright by Holt, Rinehart and Winston. All rights reserved. 4. EALUATE ADDITIONAL PRACTICE From the value for x the wrench s final speed can be determined as 93.8 s, or nearly 340 kh.. Suppose a safety net at one of the floors of the International Financial Center catches the wrench in Problem 2F. The wrench falls into the net with a velocity of 49.5 s downward. How far above the ground is the safety net located? 2. A gumdrop is released from rest at the top of the Empire State Building, which is 38 m tall. Disregarding air resistance, calculate the displacement of the gumdrop after.00, 2.00, and 3.00 s. 3. A small sandbag is dropped from rest from a hovering hot-air balloon. After 2.0 s, how far below the balloon is the sand bag? Problem F Ch. 2

2 NAME DATE CLASS 4. A physics student throws a softball straight up into the air with a speed of 7.5 s. The ball is in the air for a total of 3.60 s before it is caught at its original position. How high does the ball rise? 5. A surface probe lands on a highland region of the planet Mercury. A few hours later the ground beneath the probe gives way and the probe falls, landing below its original position with a velocity of.2 s downward. If the free-fall acceleration near Mercury s surface is 3.70 s 2 downward, what is the probe s displacement? 6. A ball thrown vertically is caught by the thrower after 5. s. Find the maximum height the ball reaches. 7. Find the initial velocity with which the ball in problem 6 is thrown. 8. An archer fires an arrow directly upward, then quickly runs from the launching spot to avoid being struck by the returning arrow. If the arrow s initial velocity is 85. s upward how long does the archer have to run away before the arrow lands? 9. A popular scene in recent action films shows a character in free-fall speed up to catch a freely falling parachute. Suppose a packed parachute is dropped from rest from an airplane and that a daredevil is launched straight down from the plane 3.00 s later. Neglecting air resistance, the daredevil catches up to the parachute 4.00 s after the daredevil leaves the plane. What are the daredevil s initial and final velocities? 0. The elevators in the Landmark Tower in Yokohama, Japan, are among the fastest in the world. They accelerate upward at 3.25 s 2 for 4.00 s to reach their maximum speed. Suppose an empty elevator is moving upward with its maximum speed when the cable breaks, so that the elevator slows down, comes to a stop, and then begins to fall freely. What will the elevator s velocity be 0.00 s,.00 s, 2.00 s, and 3.00 s after the cable breaks? Copyright by Holt, Rinehart and Winston. All rights reserved. Ch. 2 2 Holt Physics Problem Bank

3 m v i =+53.0 kh v f = 0 kh a = v f 2 2 v i = 2 x [(0 kh 2 (53.0 kh 2 3 ] 0 m km (2 (42.0 m ( km 2 /h 2 3 a = s 0 m km = 2.5 s 2 (84.0 m 6. v i = 50.0 kh forward =+50.0 kh v f = 0 kh a = 9.20 s 2 backward = 9.20 s 2 v f 2 2 v i = [(0 kh 2 (50.0 kh 2 3 ] 0 m km (2( 9.20 s 2 ( km 2 /h m km 8.4 s m = 0.5 m forward 7. a = 7.56 s m v f = v 2 i + 2 a x = v f = (0 m /s 2 + ( (7.5 6 m /s 2 ( 9. 0 m 28 7 m 2 s / 2 =±6.9 s = 6.9 s 8. v i =.8 s v f = 9.4 s a = 6. s 2 v f 2 2 v i (9.4 s 2 (.8 s 2 = = (2(6. s m /s (2 ( 6. s 2 = 7.0 m 88 m 2 /s m 2 /s 2 (2(6. s 2 9. v i =.50 s to the right =+.50 s v f = 0.30 s to the right =+0.30 s a = 0.35 s 2 to the left = 0.35 s 2 0. a = s 2 v f = 8.33 s 46.3 m Additional Practice F. v f = 49.5 s downward = 49.5 s tot = 448 m v f 2 2 v i = (0.30 s 2 (.50 s 2 (2( 0.35 s m 2 /s m 2 /s s m / s 2 =+3. m = 3. m to the right s v i = v 2 f 2 a x = (8.3 3 m /s 2 ( ( m /s 2 ( m v i = 69.4 m 2 s / m 2 s / 2 = 6. 6 m 2 s / 2 =±2.6 s = 2.6 s x i = v f 2 2 v 2 2 i ( 49.5 s 2 (0 s m /s = = (2 ( 9.8 s 2 = 25 m (2( 9.8 s 2 x 2 = x tot x = ( 448 m ( 25 m = 323 m distance from net to ground = magnitude x 2 = 323 m Section Five Problem Bank Ch. 2 7

4 2. t =.00 s t 2 = 2.00 s t 3 = 3.00 s Because, x = 2 a t 2 = 2 ( 9.8 s2 (.00 s 2 = x 2 = 2 a t 2 2 = 2 ( 9.8 s2 (2.00 s 2 = 9.6 m x 3 = 2 a t 3 2 = 2 ( 9.8 s2 (3.00 s 2 = 44. m 4.90 m 3. t = 2.0 s Because, 2 a t2 = 2 ( 9.8 s2 (2.0 s 2 = m distance of bag below balloon = m 4. v i =+7.5 s v f = 0.0 s t tot = 3.60 s x tot = 0 m 2 (v i + v f t t top = t tot = 3.6 0s =.80 s 2 2 (7.5 s s(.80 s = 5.8 m 2 5. v f =.4 s downward =.4 s a = 3.70 s 2 downward = 3.70 s 2 v f 2 2 v i (.4 s 2 (0 s m 2 /s 2 = = = 7.6 m 7.40 s 2 (2( 3.70 s m downward 6. t tot = 5.0 s t down = 2 t top To find the ball s maximum height, calculate the displacement from that height to its original position. The time interval for this free-fall is 2 t tot, and. x down = 2 a t down 2 = 2 a 2 t tot 2 = 2 ( 9.8 0s s = 3.9 m ball s maximum height = 3.9 m 7. t tot = 5.0 s x tot = 0 m 8. v i = 85. s upward =+85. s 0 m t tot = v i t tot + 2 a t tot 2 Because x tot = 0, v i = 2 a t tot = 2 ( 9.8 s2 (5.0 s =+25.0 s = 25.0 s upward Because 0 m, v i t + 2 a t 2 = 0 t = 2 v i = ( 2 ( 85. s 2 = 7.3 s a ( 9. 8 s Ch. 2 8 Holt Physics Solution Manual

5 9. t d = 4.00 s t p = t d s = 7.00 s x d = x p v i, p = 0 s x d = v i, d t d + 2 a t d 2 x p = v i, p t p + 2 a t p 2 v i, d t d + 2 a t d 2 = v i, p t p + 2 a t p 2 2 a( t p 2 t 2 d v i, d = + v i, p t p td td Because v i, p = 0 s, 2 ( 9.8 s2 [(7.00 s 2 (4.00 s 2 ] 2 a( t p 2 t 2 d v i, d = = td 4.00 s (9.8 s 2 (49.0 s s 2 (9.8 s 2 (33.0 s 2 v i, d = = = 40.5 s 8.00 s 8.00 s v f, d = a t d + v i, d = ( 9.8 s 2 (4.00 s + ( 40.5 s = ( 39.2 s + ( 40.5 s v f, d = 79.7 s 0. a up = 3.25 s 2, upward =+3.25 s 2 t up = 4.00 s v f = a up t up + v i = (3.25 s 2 (4.00 s + 0 s = 2.5 s When the cable brakes, the upward-moving elevator undergoes free-fall acceleration. v i =+2.5 s t = 0.00 s t 2 =.00 s t 3 = 2.00 s t 4 = 3.00 s v f, = a t + v i = ( 9.8 s 2 (0.00 s s = +2.5 s v f, 2 = a t 2 + v i = ( 9.8 s 2 (.00 s s = 9.8 s s = +2.7 s v f, 3 = a t 3 + v i = ( 9.8 s 2 (2.00 s s = 9.6 s s = 7. s v f, 4 = a t 4 + v i = ( 9.8 s 2 (3.00 s s = 29.4 s s = 6.9 s Section Five Problem Bank Ch. 2 9