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t= 2 ( ∆y ) a 2 ( −23 m )
= 2.2 s .
− 9.80 m s 2 = ( ) (b) The final velocity is v f = 0 + − 9.80 m s 2 m s 2 ( 2 .2 s ) = − 22 m s .
(c) The time take for the sound of the impact to reach the spectator is t sound = ∆y v sound = 23 m
= 6.8 × 10−2 s ,
340 m s so the total elapsed time is t total = 2 .2 s + 6.8 × 10−2 s ≈ 2.3 s
2.65 (a) Since the sound has constant velocity, the distance it traveled is
3
∆x = v sound t = (1100 ft s ) ( 5.0 s ) = 5.5 × 10 ft . (b) The plane travels this distance in a time of 5.0 s + 10 s = 15 s , so its velocity must be
∆x 5.5 × 103 ft
v plane =
=
= 3.7 × 102 ft s .
t
15 s 46 CHAPTER 2 (c) The time the light took to reach the observer was
t light = 5.5 × 103 ft ⎛ 1 m s ⎞
∆x
−6
=
8
⎜ 3.281 ft s ⎟ = 5.6 × 10 s .
v light 3.00 × 10 m s ⎝
⎠ D u r in g t h is t im e t h e p la n e w o u ld o n ly t r a v e l a d ist a n ce o f 0.002 ft . 2.66 The total time for the safe to reach the ground is found from 2 ( ∆y )
2 ( −25.0 m )
1
=
= 2.26 s .
∆y = v i t + at 2 with v i = 0 as t total =
2
a
−9.80 m s 2
The time to fall the first fifteen meters is found similarly: t15 = 2 ( ∆y ) a = 2 ( −15.0 m )
= 1.75 s .
−9.80 m s 2 The time Wile E. Coyote has to reach safety is
∆t = t total − t15 = 2 .26 s − 1.75 s = 0.509 s . 2.67 The time required for the woman to fall 3.00 m, starting from rest, is found from
1
1
∆y = v i t + at 2 as −3.00 m = 0 + −9.80 m s 2 t 2 , giving t = 0.782 s .
2
2 ( ) (a) With the horse moving with constant velocity of 10.0 m s , the horizontal distance is
∆x = v horse t = (10.0 m s ) ( 0.782 s ) = 7.82 m . (b) The required time is t = 0.782 s as calculated above. 47 CHAPTER 2.68 2 Assume that the ball falls 1.5 m, from rest, before touching the ground. Further, assume
that after contact with the ground the ball moves with constant acceleration for an
additional 0.50 cm (hence compressing the ball) before coming to rest.
With the first assumption, v 2 = v i2 + 2a( ∆y ) gives the velocity of the ball when...
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 Spring '05
 Dr.Ha
 Physics, Acceleration, Velocity

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