Unformatted text preview: = 35 m/s. The final velocity is
2 v = v0 + at = 35 m/s + ( 9.8 m/s )(5.0 s) = − 14 m/s (2.4) Instantaneous speed is the magnitude of the instantaneous velocity, so we drop the minus
sign and find that
Speed = 14 m/s 45. REASONING AND SOLUTION The figure at the right
shows the paths taken by the pellets fired from gun A and
gun B. The two paths differ by the extra distance covered
by the pellet from gun A as it rises and falls back to the
edge of the cliff. When it falls back to the edge of the
cliff, the pellet from gun A will have the same speed as
the pellet fired from gun B, as Conceptual Example 15
discusses. Therefore, the flight time of pellet A will be
greater than that of B by the amount of time that it takes
for pellet A to cover the extra distance. A
____
B The time required for pellet A to return to the cliff edge after being fired can be found from
Equation 2.4: v = v0 + a t. If "up" is taken as the positive direction then v0 = +30.0 m/s and
v = –30.0 m/s. Solving Equation 2.4 for t gives
t= v − v0
a = (−30.0 m/s) − (+30.0 m/s)
=
−9.80 m/s2 6.12 s Notice that this result is independent of the height of the cliff.
______________________________________________________________________________
46. REASONING
Assuming that air resistance can be neglected, the acceleration is the same for both the
upward and downward parts, namely −9.80 m/s2 (upward is the positive direction).
Moreover, the displacement is y = 0 m, since the final and initial positions of the ball are the 70 KINEMATICS IN ONE DIMENSION same. The time is given as t = 8.0 s. ( y = v0 t + 1 a t2
2 Therefore, we may use Equation 2.8 ) to find the initial velocity v0 of the ball.
( ) SOLUTION Solving Equation 2.8 y = v0 t + 1 a t 2 for the initial velocity v0 gives
2 v0 = y − 1 a t2
2 ( ) 0 m − 1 −9.80 m/s2 ( 8.0 s )
2 2 =
= +39 m/s
t
8.0 s
______________________________________________________________________________
47. REASONING AND SOLUTION In a time t the card will undergo a vertical displacement
y given by
y = 1 at2
2
where a = –9.80 m/s2. When t = 60.0 ms = 6.0 × 10–2 s, the displacement of the card is
0.018 m, and the distance is the magnitude of this value or d1 = 0.018 m .
Similarly, when t = 120 ms, d 2 = 0.071 m , and when t = 180 ms, d3 = 0.16 m .
______________________________________________________________________________
48. REASONING The minimum time that a player must wait before touching the basketball is
the time required for the ball to reach its maximum height. The initial and final velocities
are known, as well as the acceleration due to gravity, so Equation 2.4 ( v = v0 + a t ) can be
used to find the time.
SOLUTION Solving Equation 2.4 for the time yields v − v0 0 m / s − 4.6 m / s
= 0.47 s
a
−9.8 m / s 2
______________________________________________________________________________ t= 49. = SSM REASONING AND SOLUTION Equation 2.8 can be used to determine the
displacement that the ball covers as it falls halfway to the ground. Since the ball falls from
rest, its initial velocity is zero. Taking down to be the negative direction, we have y = v0t + 1 at 2 = 1 at 2 = 1 (–9.80 m/s 2 )(1.2 s) 2 = –7.1 m
2
2
2 Chapter 2 Problems 71 In falling all the way to the ground, the ball has a displacement of y = –14.2 m . Solving
Equation 2.8 with this displacement then yields the time
2y
2(–14.2 m)
=
= 1.7 s
a
–9.80 m/s 2
______________________________________________________________________________
t= 50. REASONING The initial velocity of the compass is +2.50 m/s. The initial position of the
compass is 3.00 m and its final position is 0 m when it strikes the ground. The displacement
of the compass is the final position minus the initial position, or y = –3.00 m. As the
compass falls to the ground, its acceleration is the acceleration due to gravity,
a = –9.80 m/s2. Equation 2.8 y = v0t + 1 a t 2 can be used to find how much time elapses ( 2 ) before the compass hits the ground.
SOLUTION Starting with Equation 2.8, we use the quadratic equation to find the elapsed
time. t= ( 1 a ) ( − y ) = − ( 2.50 m/s ) ± ( 2.50 m/s )2 − 4 ( −4.90 m/s2 ) − ( −3.00 m ) 2
1a
2
2( 2 )
2 ( −4.90 m/s ) 2
−v0 ± v0 − 4 There are two solutions to this quadratic equation, t1 = 1.08 s and t2 = –0.568 s. The
second solution, being a negative time, is discarded.
______________________________________________________________________________
51. REASONING AND SOLUTION
a. 2
v 2 = v0 + 2ay ( ) v = ± (1.8 m/s ) + 2 –9.80 m/s 2 ( –3.0 m ) = ±7.9 m/s
2 The minus is chosen, since the diver is now moving down. Hence, v = −7.9 m/s .
b. The diver's velocity is zero at his highest point. The position of the diver relative to the
board is
y=– 2
v0 2a =– (1.8 m/s )2 ( 2 –9.80 m/s 2 ) = 0.17 m The position above the water is 3.0 m + 0.17 m = 3.2 m .
______________________________________________________________________________ 72 KINEMATICS IN ONE DIMENSION ( ) 2
Equation 2.9 v 2 = v0 + 2ay can be used to find out how far above the 52. REASONING cliff's edge...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details