Unformatted text preview: the bullet after it enters the building is, taking down as the
negative direction, equal to –0.50 m. Therefore, the vertical component of the velocity of
the bullet as it passes through the window is, from Equation 3.5b,
v0 y (window) = 1
2 y − a yt 2
t = y1
–0.50 m 1
− a yt =
− (−9.80 m/s 2 )(0.0203 s) = –24.5 m/s
2
t
0.0203 s 2 The vertical displacement of the bullet as it travels between the buildings is (according to
Equation 3.6b with v0 y = 0 m/s) y= v2
y
2a y = (−24.5 m/s)2
= −30.6 m
2(–9.80 m/s 2 ) Therefore, the distance H is
H = 30.6 m + 0.50 m = 31 m
The time for the bullet to reach the window, according to Equation 3.4b, is
t1 = 2y
2 y 2(–30.6 m)
=
=
= 2.50 s
v0 y + v y v y (–24.5 m/s) 134 KINEMATICS IN TWO DIMENSIONS Hence, the distance D is given by D = v0 xt1 = (340 m/s)(2.50 s) = 850 m 50. REASONING AND SOLUTION Let H be the initial height of the can above the muzzle of
the rifle. Relative to its initial position the vertical coordinate of the bullet at time, t, is
y = v0 y t − 1 gt 2
2 Relative to the can’s initial position the vertical coordinate of the can at the same time is
2
y' = (1/2)gt .
NOTE: y = H – y' if the bullet hits the can. Then y = v0yt – y'. The horizontal distance
traveled by the bullet in time t is x = v0xt.
Solving for t and substituting gives
y = (v0y/v0x)x – y'
Now v0y/v0x = tan θ and it is seen from the figure in the text that H = x tan θ if the bullet is
to hit the can so y = H – y'. Hence, both objects are at the same place at the same time, and
the bullet will always strike the can. 51. REASONING The drawing shows the
trajectory of the ball, along with its
initial speed v0, horizontal displacement
x, and vertical displacement y. The angle
that the initial velocity of the ball makes
with the horizontal is θ. The known data
are shown in the tables below: +y
+x v0
y θ
x xDirection Data x ax +26.9 m 0 m/s vx
2 v0x
+(19.8 m/s) cos θ t Chapter 3 Problems 135 yDirection Data
y ay +2.74 m −9.80 m/s2 vy t v0y
+(19.8 m/s) sin θ There are only two known variables in each table, so we cannot directly use the equations of
kinematics to find the angle θ. Our approach will be to first use the x direction data and
obtain an expression for the time of flight t in terms of x and v0x. We will then enter this
expression for t into the ydirection data table. The four variables in this table, y, ay, v0y, and
t, can be related by using the appropriate equation of kinematics. This equation can then be
solved for the angle θ.
SOLUTION Using the xdirection data, Equation 3.5a can be employed to find the time t
that the ball is in the air: ( since a x = v0 x t + 1 ax t 2 = v0 x t
2 Solving for t gives t= x = 0 m/s 2 ) x
+26.9 m
=
v0 x + (19.8 m/s ) cos θ Using the expression above for the time t and the data in the ydirection data table, the
displacement in the y direction can be written with the aid of Equation 3.5b:
y = v0 y t + 1 a y t 2
2 +26.9 m
+
+2.74 m = 19.8 m/s sin θ 19.8 m/s cos θ 144 2444 4
3 ( ) ( ) =t 2 + 6.9
1
( −9.80 m/s ) (19.82m/s )m θ 2 cos 144 2444
4
3
2 = t2 Evaluating the numerical factors and using the fact that sin θ /cos θ = tan θ , the equation
above becomes
( −9.04 m )
+2.74 m = ( +26.9 m ) tan θ +
cos 2 θ
1
Using
= 1 + tan 2 θ , this equation can be rearranged and placed into a quadratic form:
2
cos θ ( −9.04 m ) tan 2 θ + ( 26.9 m ) tan θ − 11.8 m = 0
The solutions to this quadratic equation are 136 KINEMATICS IN TWO DIMENSIONS tan θ = −26.9 m ± ( 26.9 m ) − 4 ( −9.04 m )( −11.8 m )
=
2 ( −9.04 m )
2 0.535 and 2.44 The two angles are θ1 = tan −1 0.535 = 28.1° and θ 2 = tan −1 2.44 = 67.7°
______________________________________________________________________________
52. REASONING Since car A is moving faster, it will eventually catch up with car B. Each car
is traveling at a constant velocity, so the time t it takes for A to catch up with B is equal to
the displacement between the two cars (x = +186 m) divided by the velocity vAB of A
relative to B. (If the relative velocity were zero, A would never catch up with B). We can
find the velocity of A relative to B by using the subscripting technique developed in Section
3.4 of the text.
vAB = velocity of car A relative to car B
vAG = velocity of car A relative to the Ground = +24.4 m/s
vBG = velocity of car B relative to the Ground = +18.6 m/s
We have chosen the positive direction for the displacement and velocities to be the direction
in which the cars are moving. The velocities are related by
vAB = vAG + vGB
SOLUTION The velocity of car A relative to car B is
vAB = vAG + vGB = +24.4 m/s + (−18.6 m/s) = +5.8 m/s,
where we have used the fact that vGB = − vBG = −18.6 m/s. The time it takes for car A to
catch car B is
x
+186 m
t=
=
= 32.1 s
v AB +5.8 m/s
______________________________________________________________________________ Chapter 3 Problems 53. SSM REASONING The velocity vSG of the
swimmer relative to the ground is the ve...
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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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