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# 0 m s 2 2 00 m s 2 150 m 557 m s 2 after

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Unformatted text preview: with constant velocity, undergoes a displacement of ( ∆y )copter = vi t + 2 at 2 = ( −1.5 m s ) ( 2.00 s ) + 0 = − 3.00 m . 1 During this 2 .00 s , both the mailbag and the helicopter are moving downward. At the end, the mailbag is 22 .6 m − 3.00 m = 19.6 m below the helicopter. 36 CHAPTER 2 (c) Here, ( v i )bag = ( v i ) copt er = + 1.50 m s and abag = −9.80 m s 2 while acopter = 0 . After 2.00 s , the speed of the mailbag is (v ) f bag = 1.50 m m⎛ m⎞ m + ⎜ −9.80 2 ⎟ ( 2.00 s ) = − 18.1 = 18.1 . s s⎝ s⎠ s In this case, the helicopter rises 3.00 m during the 2.00 s interval while the mailbag has a displacement of −18.1 m s + 1.50 m s ⎤ ⎥ ( 2 .00 s ) = −16.6 m 2 ⎣ ⎦ ( ∆y )bag = ⎡ ⎢ from the release point. Thus, the separation between the two at the end of 2 .00 s is 3.00 m − ( − 16.6 m ) = 19.6 m . 2.48 1 (a) Consider the relation ∆y = v i t + at 2 with a = − g . When the ball is at the throwers 2 1 hand, the displacement ∆y is zero, or 0 = v i t − gt 2 . This equation has two 2 solutions, t = 0 which corresponds to when the ball was thrown, and t = 2v i g corresponding to when the ball is caught. Therefore, if the ball is caught at t = 2 .00 s , the initial velocity must have been ( ) 9.80 m s 2 ( 2 .00 s ) gt vi = = = 9.80 m s . 2 2 (b) From v 2 = v i2 + 2 a( ∆y ) , with v f = 0 at the maximum height, f ( ∆y )max = v 2 − v i2 f 2a = 0 − ( 9.80 m s ) ( 2 2 − 9.80 m s 2 ) = 4.90 m . 37 CHAPTER 2.49 2 (a) When it reaches a height of 150 m, the speed of the rocket is v f = v i2 + 2a( ∆y ) = ( 50.0 ( ) m s ) + 2 2 .00 m s 2 (150 m ) = 55.7 m s . 2 After the engines stop, the rocket continues moving upward with an initial velocity of v i = 55.7 m s and acceleration a = − g = −9.80 m s 2 . When the rocket reaches maximum height, v f = 0 . The displacement of the rocket above the point where the engines stopped (i.e., above the 150 m level) is ∆y = v 2 − v i2 f 2a = 0 − ( 55.7 m s ) ( 2 − 9.80 m s 2 2 ) = 158...
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