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# 0 m to find 2 t 2 y 2 760 m 980 m s 2 394 s

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Unformatted text preview: = tr − 0 + i . 2a v 0 v0 v0 (b) With si = 16 m , v 0 = 60 k m h , a = − 2.0 m s 2 , and t r = 1.1 s , t light = 1.1 s − 2.43 ⎛ 0.278 m s ⎞ 16 m ⎛ 1 k m h ⎞ ⎜ 1 k m h ⎠ + 60 k m h ⎝ 0.278 m s ⎠ = 6.2 s . ⎟ ⎜ ⎟ 2 − 2 .0 m s ⎝ ( 60 k m h 2 ) (a) From v 2 = v i2 + 2a( ∆y ) with v f = 0 , we have f ( ∆y )max = v 2 − v i2 f 2a = 0 − ( 25.0 m s ) ( 2 −9.80 m s 2 2 ) = 31.9 m . (b) The time to reach the highest point is t up = v f − vi a = 0 − 25.0 m s = 2.55 s . − 9.80 m s 2 (c) The time required for the ball to fall 31.9 m, starting from rest, is found from 2 ( ∆y ) 2 ( − 39.1 m ) 1 = = 2.55 s . ∆y = ( 0) t + at 2 as t = − 9.80 m s 2 a 2 (d) The velocity of the ball when it returns to the original level (2.55 s after it starts to fall from rest) is ( ) v f = v i + at = 0 + − 9.80 m s 2 ( 2 .55 s ) = − 25.0 m s . 35 CHAPTER 2.44 From v 2 = v i2 + 2a( ∆y ) , with v i = 0, v f = 29 000 km h , a n d ∆y = + 18 m , f ( ⎡ 2.9 × 10 4 k m h a= =⎣ 2 ( ∆y ) 2 (18 m ) v 2 − v i2 f 2.45 2 ) 2 − 0 ⎤ ⎛ 0.278 m s ⎞ 2 6 2 ⎦ ⎜ 1 k m h ⎟ = 1.8 × 10 m s . ⎝ ⎠ Assume the whales are traveling straight upward as they leave the water. Then v 2 = v i2 + 2 a( ∆y ) , with v f = 0 w h en ∆y = + 7.5 m , gives f ( ) v i = v 2 − 2 a( ∆y ) = 0 − 2 −9.8 m s 2 ( 7.5 m ) = 12 m s . f 2.46 1 Use ∆y = v i t + at 2 , w ith v i = 0, a = −9.80 m s 2 , a n d ∆y = −76.0 m to find 2 t= 2.47 2 ( ∆y ) a = 2 ( − 76.0 m ) −9.80 m s 2 = 3.94 s . (a) After 2 .00 s , the velocity of the mailbag is (v ) f bag ( ) = v i + at = −1.50 m s + −9.80 m s 2 ( 2 .00 s ) = − 21.1 m s . The negative sign tells that the bag is moving downward and the magnitude of the velocity gives the speed as 21.1 m s (b) The displacement of the mailbag after 2 .00 s is ⎡ −21.1 m s + ( −1.50 m s ) ⎤ ⎛ v f + vi ⎞ t=⎢ ⎥ ( 2.00 s ) = − 22.6 m . 2⎟ 2 ⎠ ⎢ ⎥ ⎣ ⎦ ( ∆y )bag = ⎜ ⎝ During this time, the helicopter, moving downward...
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