This preview shows page 1. Sign up to view the full content.
Unformatted text preview: m/s ) t + 1 a yt 2 = 1 a yt 2
2
2
SOLUTION Substituting Equation (4) into Equation (3) gives v0 = ( ∆r )2 − y 2
t = ( ∆r )2 − 1 a 2t 4
4y
t Thus, the car’s speed just as it drives off the end of the dock is ( 7.0 m )2 − 1 ( −9.80 m/s2 ) (1.1 s )4
4
2 v0 = 1.1 s = 3.4 m/s (4) 124 KINEMATICS IN TWO DIMENSIONS 39. REASONING The speed v of the soccer ball just before the goalie catches it is given by
2
v = v x + v 2 , where vx and vx are the x and y components of the final velocity of the ball.
y The data for this problem are (the +x direction is from the kicker to the goalie, and the +y
direction is the “up” direction):
xDirection Data
x ax +16.8 m 0 m/s t vx
? 2 v0x
+(16.0 m/s) cos 28.0° = +14.1 m/s
yDirection Data y ay
−9.80 m/s t vy
2 v0y ? +(16.0 m/s) sin 28.0° = +7.51 m/s
2 Since there is no acceleration in the x direction (ax = 0 m/s ), vx remains the same as v0x, so
vx = v0x = +14.1 m/s. The time t that the soccer ball is in the air can be found from the
xdirection data, since three of the variables are known. With this value for the time and the
ydirection data, the y component of the final velocity can be determined.
SOLUTION Since ax = 0 m/s2, the time can be calculated from Equation 3.5a as
x
+16.8 m
t=
=
= 1.19 s . The value for vy can now be found by using Equation 3.3b
v0 x +14.1 m/s
with this value of the time and the ydirection data:
v y = v0 y + a y t = +7.51 m/s + ( −9.80 m/s 2 ) (1.19 s ) = −4.15 m/s The speed of the ball just as it reaches the goalie is
2
v = vx + v 2 =
y ( +14.1 m/s ) + ( −4.15 m/s )
2 2 = 14.7 m/s ______________________________________________________________________________
40. REASONING AND SOLUTION On impact
vx = v cos 75.0° = (8.90 m/s) cos 75.0° = 2.30 m/s
and
2 2 2 2 2 v0y = vy + 2gy = (8.90 m/s) sin 75.0° + 2(9.80 m/s )(–3.00 m)
so v0y = 3.89 m/s Chapter 3 Problems 125 The magnitude of the diver's initial velocity is
2
2
v0 = ( 2.30 m/s ) + ( 3.89 m/s ) = 4.52 m/s The angle the initial velocity vector makes with the horizontal is θ0 = tan−1 (voy /vox) = 59.4°
_____________________________________________________________________________________________ 41. REASONING As discussed in Conceptual Example 5, the horizontal velocity component
of the bullet does not change from its initial value and is equal to the horizontal velocity of
the car. The same thing is true here for the tomato. In other words, regardless of its vertical
position relative to the ground, the tomato always remains above you as you travel in the
convertible. From the symmetry of free fall motion, we know that when you catch the
tomato, its velocity will be 11 m/s straight downward. The time t required to catch the
tomato can be found by solving Equation 3.3b ( v y = v0 y + a y t ) with v y = – v0 y . Once t is
known, the distance that the car moved can be found from x = vxt .
SOLUTION Taking upward as the positive direction, we find the flight time of the tomato
to be
v y – v0 y –2v0 y –2(11 m/s)
t=
=
=
= 2.24 s
ay
ay
–9.80 m/s 2
Thus, the car moves through a distance of x = vx t = (25 m/s )(2.24 s) = 56 m 42. REASONING As shown in the drawing, the angle
that the velocity vector makes with the horizontal is
given by
vy
tan θ =
vx y
v θ
vx where, from Equation 3.3b,
v y = v0 y + a y t = v0 sin θ 0 + a y t and, from Equation 3.3a (since ax = 0 m/s2), v x = v0 x = v0 cos θ 0 vy
x 126 KINEMATICS IN TWO DIMENSIONS Therefore, tan θ = vy
vx = v0 sin θ 0 + a y t
v0 cos θ0 SOLUTION Solving for t, we find
t= v0 ( cos θ 0 tan θ – sin θ0 )
ay = ( 29 m/s ) ( cos 36° tan 18° – sin 36° ) =
–9.80 m/s 2 0.96 s ____________________________________________________________________________________________ 43. SSM WWW REASONING The horizontal distance covered by stone 1 is equal to the
distance covered by stone 2 after it passes point P in the following diagram. Thus, the
distance ∆x between the points where the stones strike the ground is equal to x2, the
horizontal distance covered by stone 2 when it reaches P. In the diagram, we assume up and
to the right are positive.
x
2 2
θ P
θ θ 1 ∆x SOLUTION If tP is the time required for stone 2 to reach P, then
x2 = v0 xtP = (v0 cos θ )tP For the vertical motion of stone 2, v y = v0 sin θ + a y t . Solving for t gives
t= v y − v0 sin θ
ay Chapter 3 Problems 127 When stone 2 reaches P, v y = −v0 sin θ , so the time required to reach P is
tP = −2v0 sin θ
ay Then, −2v sin θ
0
x2 = v0 x tP = (v0 cos θ ) ay x2 = 2
−2v0 sin θ cos θ ay = −2(13.0 m/s)2 sin 30.0° cos 30.0°
= 14.9 m
–9.80 m/s 2 v0x
Since the vertical
44. REASONING
vx
height is asked for, we will begin
with the vertical part of the motion,
65°
treating it separately from the
horizontal part.
The directions
vy
H
upward and to the right are chosen as
the positive directions in the drawing.
The data for the vertical motion are
vx
summarized in the following table.
Note that the initial velocity
component v0y is zero, because the
vy
marble is thrown horizontally. The
vertical c...
View
Full
Document
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

Click to edit the document details