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and find that
1
2 x = v0 xt + a xt 2 or a x =
ax = ( 2 x − 2v0 xt
t 2 (3.5a) ) 2 4.11×106 m − 2 ( 4370 m/s )( 684 s ) ( 684 s ) 1
2 y = v0 y t + a y t 2 or a y =
ay = 2 = 4.79 m/s 2 ( 2 y − 2v0 y t
t2 (3.5b) ) 2 6.07 × 106 m − 2 ( 6280 m/s )( 684 s ) ( 684 s )2 = 7.59 m/s 2 67. REASONING The vertical component of the ball’s velocity v0 changes as the ball
approaches the opposing player. It changes due to the acceleration of gravity. However, the
horizontal component does not change, assuming that air resistance can be neglected.
Hence, the horizontal component of the ball’s velocity when the opposing player fields the
ball is the same as it was initially.
SOLUTION Using trigonometry, we find that the horizontal component is b g v x = v 0 cosθ = 15 m / s cos 55° = 8 . 6 m / s 68. REASONING Since the spider encounters no appreciable air resistance during its leap, it
can be treated as a projectile. The thickness of the magazine is equal to the spider’s vertical
displacement y during the leap. The relevant data are as follows (assuming upward to be the
+y direction):
yDirection Data
y ay ? −9.80 m/s2 vy v0y t (0.870 m/s) sin 35.0° = +0.499 m/s 0.0
770 s SOLUTION We will calculate the spider’s vertical displacement directly from y = v0 yt + 1 a yt 2 (Equation 3.5b):
2 ( ) y = ( 0.499 m/s )( 0.0770 s ) + 1 −9.80 m/s2 ( 0.0770 s ) = 0.0094 m
2
2 To express this result in millimeters, we use the fact that 1 m equals 1000 mm: 148 KINEMATICS IN TWO DIMENSIONS 1000 mm Thickness = 0.0094 m = 9.4 mm
1m ( ) 69. REASONING AND SOLUTION The components of the initial velocity are
v0x = v0 cos θ = (22 m/s) cos 40.0° = 17 m/s
v0y = v0 sin θ = (22 m/s) sin 40.0° = 14 m/s
a. Solving Equation 3.6b for y gives
y= 2
v2 − v0 y
y 2a y
When the football is at the maximum height y = H, and the football is momentarily at rest,
so vy = 0 m/s. Thus,
H= 2
0 − v0 y = 2a y 0 − (14 m/s ) 2 = 6.0 × 101 m 2 ( −1.62 m/s 2 ) b. When the ball strikes the ground, y = 0 m; therefore, the time of flight can be determined
from Equation 3.5b with y = 0 m.
1 y = v0yt + 2 ayt2
or
0 = [(14 m/s) + 1
2 ( −1.62 m / s 2 ) t] t t = 17 s The range is
x = R = v0xt = (17 m/s)(17 s) = 290 m ____________________________________________________________________________________________ 70. REASONING When the skier leaves the ramp, she exhibits projectile motion. Since we
know the maximum height attained by the skier, we can find her launch speed v0 using
2
Equation 3.6b , v 2 = v0 y + 2a y y , where v0 y = v0 sin 63° .
y SOLUTION At the highest point in her trajectory, vy = 0. Solving Equation 3.6b for v0y we
obtain, taking upward as the positive direction, v 0 y = v 0 sin 63° = –2 a y y or v0 = –2 a y y
sin 63° = –2(–9.80 m / s 2 ) (13 m)
= 18 m / s
sin 63° ____________________________________________________________________________________________ Chapter 3 Problems 149 2
2
71. SSM REASONING The speed of the fish at any time t is given by v = v x + v y , where vx and vy are the x and y components of the velocity at that instant. Since the horizontal
motion of the fish has zero acceleration, vx = v0 x for all times t. Since the fish is dropped by
the eagle, v0x is equal to the horizontal speed of the eagle and v0 y = 0 . The y component of
the velocity of the fish for any time t is given by Equation 3.3b with v0 y = 0 . Thus, the
2
speed at any time t is given by v = v 0 x + ( a y t ) 2 . SOLUTION
2
2
2
2
a. The initial speed of the fish is v 0 = v 0 x + v 0 y = v 0 x + 0 = v 0 x . When the fish's speed doubles, v = 2v0 x . Therefore,
2
2v0 x = v0 x + ( a y t ) 2 or 2
2
4v0 x = v0 x + (a y t ) 2 Assuming that downward is positive and solving for t, we have
t= 3 v0 x
ay =3 F6.0 m / s I =
G m/s J
H
K
9.80
2 1.1 s b. When the fish's speed doubles again, v = 4v0 x . Therefore,
2
4v0 x = v0 x + (a y t ) 2 Solving for t, we have
t = 15 v0 x
ay = 15 or 2
2
16 v 0 x = v 0 x + ( a y t ) 2 F6.0 m / s I = 2.37 s
G m/s J
H
K
9.80
2 Therefore, the additional time for the speed to double again is ( 2 .4 s) – (1.1 s) = 1.3 s .
____________________________________________________________________________________________ 72. REASONING We will treat the horizontal and vertical parts
v0x
of the motion separately. The directions upward and to the
right are chosen as the positive directions in the drawing.
y
Ignoring air resistance, we note that there is no acceleration
in the horizontal direction. Thus, the horizontal component
x
v0x of the ball’s initial velocity remains unchanged
throughout the motion, and the horizontal component x of the displacement is simply v0x
times the time t during which the motion occurs. This is true for both bunted balls. Since the
value of x is the same for both, we have that 150 KINEMATICS IN TWO DIMENSIONS t (1)
v0 x,B = v0 x,A A t B
To use this result to calculate v0x,B, it is necessary to determine the times tA and tB. We can
accomplish this by applying the equations of kinematics to the vertical part of the motion for...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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