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**Example text**

0 m s . 48. 0 s. Use Eq. (2-15). 0 s)2 = 13 m . 0 s) 49. Strategy Use Eq. (2-15) to find the time it takes for the coin to reach the water. Then, find the time it takes the sound to reach Glenda’s ear. Add these two times. 00 m. Solution Find the time elapsed between the release of the coin and the hearing of the splash. 1 1 2h h h = viy ∆t + a y (∆t ) 2 = 0 + g (∆t1 ) 2 , so ∆t1 = . h = vs ∆t2 , so ∆t2 = . 22 s . 80 m s 50. (a) Strategy The stone is instantaneously at rest at its maximum height.

0 km 2 2 (b) Strategy viy = the speed when the engine fails = a∆t1; v y = viy − g ∆t = 0 at maximum height. Solution Find the time elapsed from the engine failure to maximum height. 0 s) = 102 s. 0 s = 152 s . (c) Strategy Use Eq. (2-15). Solution Find the maximum height reached by the rocket. 80 N kg) (d) Strategy Use Eq. (2-16). viy = 0 at the maximum height. Solution Find the final velocity. 0 × 103 m) = 1220 m s. G Thus, v = 1220 m s downward . 70. Strategy Analyze the graph to answer each question about the motion of the Engine.

10 h ∆t 34. (a) Strategy Find the average speed by dividing the total distance traveled by the total time. 0. 0 (b) Strategy Use the definition of average velocity. Draw a diagram. Solution Let east be the +x-direction and north be the +y-direction. N 30° 128 km 60° 96 km G Find the magnitude of ∆r. 0 h) 2 = 195 km G Find the direction of ∆r. 65 km ∆r G G = = 97 km h and v av = = 97 km h at 35° north of east . 0 h ∆t 222 Physics Chapter 3: Motion in a Plane 35. (a) Strategy Find the average speed by dividing the total distance traveled by the total time.