College Physics Kinematics Solution Manual Part 2

OpenStax College Physics 31. Instructor Solutions Manual A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0 .350 m/s airborne? (b) How long does this take? Solution 2 (a) (b) 32. Chapter 2 x−x 0 = t= 2 , how far will it travel before becoming 2 v −v 0 ( 6. 00 m/s )2 −( 0 m/s )2 =51 . 4 m = 2a 2(0 .350 m/s 2 ) v − v 0 6 . 00 m/s−0 m/s =17 .1 s = a 0 .350 m/s 2 Professional Application A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm. (a) Find the acceleration in m/s 2 and in multiples of g ( g=9. 80 m/s2 ) . (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g ? Solution a= v 2 −v 0 2 (0 m/s)2 −(0. 600 m/s )2 =−90 . 0 m/s2 = 2( x−x 0 ) 2(2 .00×10−3 m) |a| 90 . 0 m/s2 = =9 . 18 ⇒ a= 9 .18 g . 2 (a) |g| 9 . 80 m/s 1 (v +v )t 2 0 , so that 2 (x −x0 ) 2(2 . 00×10−3 m) = =6 .67×10−3 s t= v 0 +v (0 . 600 m/s )+(0 m/s ) (b) x − x 0= v 2 −v 0 2 (0 m/s)2 −(0. 600 m/s )2 =−40 . 0 m/s2 2( x−x 0 ) 2( 4 .50×10-3 m) |a| 40 . 0 m/s2 = =4 . 08 ⇒ a=4 . 08 g 2 (c) |g| 9 . 80 m/s a= = OpenStax College Physics 33. Chapter 2 An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last? Solution a= (a) t= (b) 34. Instructor Solutions Manual v 2−v 02 (0 m/s )2 −( 7 . 50 m/s )2 =−80 . 4 m/s2 2( x−x 0 ) 2(0. 350 m) = 2 (x −x0 ) 2(0 . 350 m) =9 . 33×10−2 s = 0 m/s + 7 .50 m/s v+ v0 In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot’s speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m. Solution Knowns: x = 3 m; v = 0 m/s; v 0 = 54 m/s We want a , so we can use this equation: v 2−v 0 2 0 m/s−( 54 m/s )2 =−486 m/s2 v 2 = v 02 +2 ax ⇒ a= = 2x 2( 3 m) . Negative acceleration means that the pilot was decelerating at a rate of 486 m/s every second. 35. Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem. OpenStax College Physics Solution (a) Chapter 2 v 2 =v 02 +2 ax ⇒ v = √ 2 ax= √2(−9 . 8 m/s2 )(− 3. 0 m )=7 .7 m/s 2 (b) Instructor Solutions Manual a= 2 v −v 0 (0 m/s)2 −(7 .7 m/s )2 = =−1 .5×103 m/s 2 2x 2(0.02 m ) This is ~ 3 times the deceleration of the pilots, who were falling from thousands of meters high! 36. An express train passes through a station. It enters with an initial velocity of 22.0 2 0 .150 m/s as it goes through. The station is 210 m/s and decelerates at a rate of m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves? Solution (a) x − x 0 =v 0 t + 1 2 at 2 , and rearranging −v 0± √ v 02 +2 a( x−x 0 ) t= a −22. 0 m/s± √ (22 .0 m/s )2 +2(−0 . 150 m/s2 )(210 m) ¿ =9 .88 s −0 .150 m/s 2 2 2 (b) v =v 0 +2 a( x − x 0 ) , and rearranging v= √ v20+2 a(x−x 0 )= √ (22. 0 m/s)2 +2 (−0 . 150 m/s2 )( 210 m )=20. 6 m/s (c) Here, we use the fact that the train will leave the station when the nose is 210 m + 130 m = 340 m away from the beginning of the station. x − x 0 =v 0 t + 1 2 at 2 −v ± √v20+2 a( x − x 0 ) t= 0 a −22. 0 m/s± √ (22 .0 m/s )2 +2 (− 0 . 150 m/s2 )( 340 m ) ¿ =16 .4 s −0 .150 m/s 2 OpenStax College Physics (d) 37. Instructor Solutions Manual Chapter 2 v 2 = v 02 +2 a( x − x 0 ) v= √(22. 0 m/s)2 +2 (−0 .150 m/s 2 )( 340 m )=19 .5 m/s Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity. Solution (a) a= 2 v − v 0 145 m/s −0 m/s =32. 6 m/s2 = 4 . 45 s t 2 (b) v =v 0 +2 a( x − x 0 ) , and rearranging v= √(0 m/s )2 +2(32. 6 m/s 2 )(402 m)=162 m/s (c) v >v max because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at 2 the beginning, so it would not be accelerating at 32. 6 m/s during the last few meters, but substantially less, and the final velocity would be less than 162 m/s . 38. Solution A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial 2 velocity of 11.5 m/s and accelerates at the rate of 0 .500 m/s for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish? 2 (a) v =v 0 + at=11.5 m/s +( 0 . 500 m/s )( 7. 00 s )=15 . 0 m/s (b) Let t const be the time it takes the rider to reach the finish line without OpenStax College Physics accelerating: Now let d Instructor Solutions Manual t const= Chapter 2 x 300 m =26 .09 s=26 . 1s = v 0 11. 5 m/s be the distance traveled during the 7 seconds of acceleration. We know t= 7. 00 s so 1 d=v 0 t+ at 2 =(11 . 5 m/s )( 7 . 00 s )+ 0. 5 ( 0 . 500 m/s2 )( 7 .00 s )2 = 92 .75 m = 92. 8 m 2 ' Let t be the time it will take the rider at the constant final velocity to complete the race: t'= x−d 300 m −92. 75 m = = 13. 82 s = 13. 8 s v 15 . 0 m/s . So the total time T it will take the accelerating rider to reach the finish line is T = t + t =7 s + 13 .82 s = 20 .82 s = 20 . 8 s. Finally, let T* be the time saved. So T*= 26.09 s −20.82 s =5.27 s . (c) Let t 2 be the time it takes for rider 2 to reach the finish line. ' t 2= x 295 m =25 . 0 s; time difference =t −T = 25 . 0 s− 20 . 817 s = 4 . 2 s = 2 v '0 11. 8 m/s Therefore he finishes 4.2 s after the winner. When the other racer reaches the finish line, the winner has been traveling at 15 m/s for 4.2 seconds, so the other racer finishes x=(4 .2 m/s)(15 m/s) = 63 m 39. Solution behind the other racer. In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course? There are two parts to the race: an acceleration part and a constant speed part. First, we need to determine how long (both in distance and time) it takes the motorcycle to finish accelerating. During acceleration, a= 60 mph 4s . OpenStax College Physics Instructor Solutions Manual ( Chapter 2 ) v max =183 mph=at 1 ⇒t 1=183 mph 4 s =12. 2 s 60 mph 183 mph + 0 mph x 1=v avg t 1= ( 12.2 s)=( 91.5 mph ) 1 h (12. 2 s)=0. 31 mi 2 3600 s ( At constant velocity, ) ( x 2= 5 mi − 0.31 mi = 4 .7 mi ) . Now, we complete the calculation by determining how much time is spent on the course at max speed. x2 4 . 7 mi 3600 s =92 s, so = =0 .026 h 1h v max 183 mph t total=t 1 + t 2 =12. 2 s+ 92 s=104 s t 2= 40. (a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100m dash, what was his maximum speed for this race? Solution (a) There are two parts to the race and must be treated separately since acceleration x 1 (while x 1 + x 2=100 m . is not uniform over the race. We will divide the race into accelerating) and x 2 (with constant speed), where When the speed is constant, When accelerating, t=3.00 s t=6. 69 s , so x 2=vt=(6 . 69 s)v=100 m−x 1 1 x 1= at 2= 1 vt=(1 .50 s)v 2 2 , so . Plugging x 1 into the previous equation, we get his maximum speed: =12 .2 m/s 100 m− (1 .50 s ) v= ( 6 . 69 s ) v ⇒ v =100 m 6 . 69 s + 1 .50 s Therefore, his acceleration was v 12. 2 m/s =4 . 07 m/s2 a= = t 3 . 00 s . OpenStax College Physics Instructor Solutions Manual Chapter 2 (b) Similar to part (a), we can plug in the different values for time and total distance: 200 m− (1 .50 s ) v= ( 16 . 30 s ) v ⇒ v = 200 m =11. 2 m/s 16. 30 s + 1 .50 s 2.7 FALLING OBJECTS 41. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y 0 =0 . Solution 2 Knowns: a= acceleration due to gravity =g=− 9. 8 m/s ; y 0= 0 m; v 0 =+ 15 . 0 m/s To find displacement we use v =v 0 +at 1 y = y 0 + v 0 t+ at 2 2 , and to find velocity we use . 1 y 1 = y 0 + v 0 t 1 + at 21 2 1 ¿ 0 m +( 15 . 0 m/s )( 0 . 500 s )+ (− 9 .8 m/s 2 )( 0 . 500 s )2=6. 28 m 2 =v +at =( 15 . 0 m/s )+(− 9. 8 m/s2 )( 0 . 500 s )=10 .1 m/s v 1 0 1 (a) 1 y 2 = y 0 +v 0 t 2 + at 22 2 1 ¿ 0 m +( 15 . 0 m/s )( 1 . 00 s )+ (− 9. 8 m/s2 )( 1 .00 s )2 =10 .1 m 2 = +at =( 15 .0 m/s )+(− 9 . 8 m/s2 )( 1. 00 s )=5 . 20 m/s v v 2 0 2 (b) 1 y 3 = y 0 +v 0 t 3 + at 32 2 1 ¿ 0 m +( 15 . 0 m/s )( 1 .50 s )+ (−9 . 8 m/s2 )( 1. 50 s )2 =11.5 m 2 =v +at =( 15 .0 m/s )+(− 9 . 8 m/s2 )( 1. 50 s )=0 . 300 m/s v 3 0 3 (c) The ball is almost at the top. OpenStax College Physics Instructor Solutions Manual Chapter 2 1 y 4 = y 0 +v 0 t 4 + at 42 2 1 ¿ 0 m +( 15 . 0 m/s )( 2 . 00 s )+ (− 9. 8 m/s 2 )( 2 .00 s )2 =10. 4 m 2 =v +at =( 15 . 0 m/s )+(− 9 . 8 m/s2 )( 2. 00 s )=− 4 . 60 m/s v 4 0 4 (d) The ball has begun to drop. 42. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water. Solution 2 Knowns: a=−9 . 8 m/s ;v 0 =−14 .0 m/s; y 0= 0 m To find displacement we use v =v 0 + at 1 y = y 0 + v 0 t+ at 2 2 , and to find velocity we use . 1 y 1 = y 0 + v 0 t1 + at 12 2 1 (−9 .8 m/s 2 )(0. 500 s )2=−8 .23 m 2 2 (a) v 1 = v 0 +at 1 =(−14 . 0 m/s )+(− 9 .8 m/s )( 0 . 500 s )=−18 . 9 m/s 1 y 2 = y 0 + v 0 t 2 + at 22 2 1 2 2 ¿ 0 m+(−14 . 0 m/s )( 1 . 00 s )+ (− 9. 8 m/s )( 1 . 00 s ) =−18 .9 m 2 2 (b) v 2 =v 0 +at 2 =(−14 .0 m/s )+(− 9. 8 m/s )( 1 .00 s )=−23. 8 m/s ¿ 0 m +(−14 .0 m/s )( 0. 500 s )+ 1 y 3 = y 0 + v 0 t 3 + at 32 2 1 ¿ 0 m +(−14 . 0 m/s )( 1 .50 s )+ (− 9. 8 m/s2 )( 1 .50 s )2 =−32. 0 m 2 =v +at =(−14 . 0 m/s)+(− 9. 8 m/s2 )( 1 .50 s )=−28. 7 m/s v 3 0 3 (c) OpenStax College Physics Instructor Solutions Manual Chapter 2 1 y 4 = y 0 +v 0 t 4 + at 42 2 1 (−9 .8 m/s 2 )(2 .00 s )2=−47 . 6 m 2 2 (d) v 4 =v 0 +at 4 =(−14 . 0 m/s )+(−9 . 8 m/s )( 2. 00 s )=−33 . 6 m/s ¿ 0 m +(−14 . 0 m/s )( 2. 00 s )+ 1 y 5 = y 0 + v 0 t 5 + at 52 2 1 2 2 ¿ 0 m+(−14 . 0 m/s )( 2. 50 s )+ (− 9. 8 m/s )( 2 .50 s ) =−65 .6 m 2 2 v 5 =v 0 +at 5 =(−14 .0 m/s )+(−9 .8 m/s )( 2 .50 s )=−38 . 5 m/s (e) 43. Solution A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball? v 2= v 02 +2 a( y− y 0 )⇒v 02=v2 −2 a( y− y 0 ) v 0= √ v 2−2 a( y− y 0 )= √ (0 m/s)2 −2(−9 .80 m/s 2 )( 1 .25 m )= 4 .9 5 m/s 44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. Solution 2 (a) Knowns: a=−9 . 80 m/s ;v 0 =−1 . 40 m/s; t= 1 .8 s; y= 0 m 1 at 2 2 (b) ¿ 0 m +(−1 .40 m/s )( 1 .8 s )+ 0 .5 (−9 .80 m/s2 )( 1 .8 s )2=18 m y 0 =−y+v 0 t+ OpenStax College Physics Instructor Solutions Manual Chapter 2 45. A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation. Solution (a) Knowns: a=−9 . 80 m/s ; v0 = 13. 0 m/s; y 0 = 0 m 2 (b) At the highest point in the jump, v = 0 m/s . We can use the equation 2 2 v −v 0 y− y0= 2a because the only unknown it includes is y , which is what we want to solve for. First we substitute the knowns into the equation and then we solve for y . y− y 0 = 2 2 v 2 −v 02 v −v 0 (0 m/s )2−( 13 .0 m/s )2 −0 m=8 . 62m − y0= ⇒ y= 2 2a 2a 2(−9 . 80 m/s ) . Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result. (c) If t is the time for the dolphin to reach its peak height, then the dolphin is out of the water. v =v 0 + at ⇒ t= 46. Solution 2t is the time v−v 0 0 m/s −13 .0 m/s =1 .3265 s, and 2 t =2 .65 s = a −9 . 8 m/s2 A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air? (b) What is her highest point above the board? (c) What is her velocity when her feet hit the water? 2 (a) Knowns: y 0 =1. 80 m, y =0 m, a=−9. 80 m/s , v 0=4. 00 m/s equation 1 y − y 0 = v 0 t+ at 2 2 . Rearranging, , so we use the OpenStax College Physics Instructor Solutions Manual Chapter 2 −v 0± √ v20−2 a( y 0 − y ) t= a −4 .00 m/s± √ ( 4 .00 m/s )2 −2 (−9 .80 m/s 2 )( 1 . 80 m ) ¿ =1. 14 s −9 . 80 m/s2 2 2 (b) y− y0= v −v 0 (0 m/s)2 −( 4 .00 m/s )2 =0.816 m = 2 2a 2(−9 .80 m/s ) 2 2 (c) v =v 0 +2 a( y − y 0 ) and rearranging v=± √ v20+2a( y− y 0 )=± √(4.00 m/s)2 +2 (−9.80 m/s2 )(−1.80 m ) ¿± √ 51.28 m2 /s2 =±7.16 m/s v =−7.16 m/s . Since the diver must be moving in the negative direction, 47. (a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would it take to reach the ground if it is thrown straight down with the same speed? Solution (a) Knowns: t= 2. 35 s; y = 0 m; v 0=+ 8. 00 m/s; a=−9. 8 m/s Since we know t , equation y , v 0 , and 1 y = y 0 + v 0 t+ at 2 2 a 2 y0 and want to find , we can use the . 1 y=( 0 m )+(+ 8 .00 m/s )( 2 .35 s)+ (−9 . 80 m/s2 )(2. 35 s)2=−8 .26 m 2 , so the cliff is 8.26 m high. (b) Knowns: y = 0 m; y 0 = 8. 26 m; v 0 =−8 . 00 m/s; a =−9 .80 m/s Now we know y , y 0 , v 0 , and a 2 and want to find t , so we use the OpenStax College Physics Instructor Solutions Manual equation 1 y = y 0 + v 0 t+ at 2 2 Chapter 2 again. Rearranging, −v 0± √ v 20−4 (0 .5 a )(y 0 − y ) t= 2(0 .5 a) −(−8 .00 m/s )± √ (−8 .00 m/s )2+2(9. 80 m/s 2)( 8 .26 m−0 m ) t= (−9 .80 m/s2 ) 8.00 m/s ±15. 03 m/s ¿ −9.80 m/s2 t=0.717 s or −2. 35 s⇒ t=0 .717 s 48. A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall? Solution Knowns: y 0 =2. 20 m; y =1. 80 m; v 0 =11.0 m/s; a=−9 .80 m/s 2 −v 0± √ v20−2 a( x − x 0 ) t= a −11 . 0 m/s ± √ (11 . 0 m/s)2 + 2 (− 9. 80 m/s 2 )(0 . 40 m ) ¿ =2 .28 s −9 . 80 m/s2 49. Solution You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down? 2 Knowns: a=−9 . 80 m/s ;v 0 = 15. 0 m/s; y= 7. 00 m 1 y = v 0 t+ at 2 ⇒ 2 −v 0± √ v20+2 ay t= a −15 .0 m/s± √ (15 . 0 m/s)2 +2(− 9. 80 m/s 2 )( 7 .00 m ) −15 .0 m/s ±9 . 37 m/s ¿ = −9 . 80 m/s2 −9 . 80 m/s2 OpenStax College Physics So t1 = 0. 58 s 1.91 s 50. Solution Instructor Solutions Manual t 2= 2. 49 s and Chapter 2 , so the total time between passing the branch is . A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air? (a) (b) v 2 = v 02 +2 ay ⇒ v 0 = √ v 2−2 ay = √ 0−2(−9 . 80 m/s2 )( 2. 50 m )= 7. 00 m/s y = v avg t= ( v 2+v ) t ⇒t= v2+yv = 2(27 .00.50m/sm ) =0. 714 s 0 0 51. Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head? Solution 1 1 y=v 0 t+ at 2 = (9 . 80 m/s2 )(1. 50 s )2=11. 0 m 2 2 (a) So the rock falls 11.0 m in the 1.50 s before the hiker sees the rock. When he finally sees the rock, it is 94.0 m above his head. (b) y= √ √ 2y 1 2 = at ⇒t = a 2 2( 105 m ) =4 . 63 s 9. 80 m/s 2 , so the rock will take 4.63 s to fall 4.63 s−1.50 s= 3.13 s the full distance. Thus, the hiker will have to move out of the way before the rock strikes the hiker's location, ignoring the height of person. 52. An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground. OpenStax College Physics Solution Instructor Solutions Manual Chapter 2 (a) 1 1 x=v 0 t+ at 2 = (9 .80 m/s 2 )( 1 . 00 s )2=4. 90 m 2 2 (b) v 2 = v 02 +2 ax ⇒ v = √2( 9. 80 m/s2 )( 75 . 0 m )=38 . 3 m/s (c) First find the total time to fall: y= √ √ 2y 1 2 = at ⇒t = a 2 2(75 . 0 m) 9 .80 m/s 2 =3 . 91 s Next, we find the distance traveled up to the last 1 second of flight: 1 1 y= at 2 = (9 .80 m/s 2 )( 2 .91 s )2=41. 5 m 2 2 second will be the difference: 53. Solution , so the distance traveled in the last x= 75.0 m −41.5 m =33.5 m . There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day. 2 (a) Knowns: v 0 = 0 m/s, y 0 = 0 m, y =−250 m, a=− 9. 80 m/s . v=± √ v20+2 a( y− y 0 )=± √(0 m/s )2 +2(−9 .80 m/s 2 )(−250 m )=−70 . 0 m/s downward. (b) Let t s = the time for the sound to travel to the tourist. height of cliff 250 m =0.7463 s t s= = 335 m/s vs Let t = the total time before the tourist can react. t= t s + reaction time = 0. 7463 s + 0.300 s = 1.046 s Let t b = the time it takes the rock to reach the bottom. t b= v − v 0 −7 .00 m/s −0 m/s = =7 . 143s . g −9. 8 m/s 2 Now subtract t b− t = 7.143 s−1.046 s =6.10 s ,