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Unformatted text preview: e to gravity is the same for both balls, despite the fact that they have
different velocities. 4. (d) The acceleration of a projectile is the same at all points on the trajectory. It points
2
downward, toward the earth, and has a magnitude of 9.80 m/s . 5. (c) Since there is no acceleration in the x direction (ax = 0 m/s ) for projectile motion, the x
component vx of the velocity is constant throughout the motion. And, the acceleration due to 2 2 gravity, ay = −9.80 m/s (“downward” is the negative direction), also remains constant.
6. (c) The time for a projectile to reach the ground depends only on the y component (or
vertical component) of its variables, i.e., y, v0y, and ay. These variables are the same for both
balls. The fact that Ball 1 is moving horizontally at the top of its trajectory does not play a
role in the time it takes for it to reach the ground. 7. (a) Using y = −19.6 m, ay = −9.80 m/s , and v0y = 0 m/s, Equation 3.5b y = v0 y t + 1 a y t 2
2 2 ( can be used to calculate the time t.
8. (b) The time a projectile is in the air is equal to twice the time it takes to fall from its
maximum height. Both have the same maximum height, so the time of fall is the same.
Therefore, both projectiles are in the air for the same amount of time. ) 98 KINEMATICS IN TWO DIMENSIONS 9. (a) The time a projectile is in the air is equal to twice the time it takes to fall from its
maximum height. Projectile 1 reaches the greater height, so it spends the greater amount of
time in the air. 10. (c) The vertical component (or y component) of the final velocity depends on the y
components of the kinematic variables ( y, v0y, and ay) and the time t. These variables are the
same for both balls, so they have the same vertical component of the velocity.
11. (a) A person standing on the ground sees a car traveling at +25 m/s. When the driver of the
car looks out the window, she sees the person moving in the opposite direction with the
same speed, or with a velocity of −25 m/s.
12. (d) The velocity vBC of the bus relative to the car is given by the relation vBC = vBG + vGC =
vBG + (−vCG) = +16 m/s + (−12 m/s) = +4 m/s. 13. (d) This answer is arrived at by using the relation vPG = vPB + vBG = +2 m/s + 16 m/s =
+18 m/s.
14. (c) The velocity vPC of the passenger relative to the car is given by vPC = vPB + vBC,
according to the subscripting method discussed in Section 3.4. However, the last term on the
right of this equation is given by vBC = vBG + vGC. So, vPC = vPB + vBG + vGC =
+2 m/s + 16 m/s + (−12 m/s) = +6 m/s.
15. (b) The velocity of the jeep relative to you is zero. Thus, the horizontal component of the
tire’s velocity relative to you is also zero. Since this component of the velocity never
changes as the tire falls, the car cannot hit the tire, regardless of how close the car is to the
jeep.
16. The magnitude vAB of the velocity of car A relative to car B is vAB = 34.2 m/s. The angle
that the velocity vAB makes with respect to due east is θ = 37.9° south of east. Chapter 3 Problems 99 CHAPTER 3 KINEMATICS IN TWO DIMENSIONS
PROBLEMS
1. SSM REASONING The displacement is a vector drawn from the initial position to the
final position. The magnitude of the displacement is the shortest distance between the
positions. Note that it is only the initial and final positions that determine the
displacement. The fact that the squirrel jumps to an
intermediate position before reaching his final position
is not important. The trees are perfectly straight and
both growing perpendicular to the flat horizontal
A
ground beneath them. Thus, the distance between the
2.5 m
trees and the length of the trunk of the second tree
below the squirrel’s final landing spot form the two
perpendicular sides of a right triangle, as the drawing
shows. To this triangle, we can apply the Pythagorean
theorem and determine the magnitude A of the
1.3 m
displacement vector A. SOLUTION According to the Pythagorean theorem, we have
A= 2. (1.3 m )2 + ( 2.5 m )2 = 2.8 m REASONING AND SOLUTION The horizontal and vertical components of the plane's
2
2
velocity are related to the speed of the plane by the Pythagorean theorem: v 2 = vh + vv .
Solving for vh we have
2
v h = v 2 − v v = (245 m / s) 2 − ( 40.6 m / s) 2 = 242 m / s 3. REASONING To determine the horizontal and vertical components of the launch velocity,
we will use trigonometry. To do so, however, we need to know both the launch angle and
the magnitude of the launch velocity. The launch angle is given. The magnitude of the
launch velocity can be determined from the given acceleration and the definition of
acceleration given in Equation 3.2. 100 KINEMATICS IN TWO DIMENSIONS SOLUTION According to Equation 3.2, we have
a= v − v0 or t − t0 340 m / s 2 = v−0 m/s
0.050 s or c v = 340 m / s 2 b
h0.050 sg Using trigonometry, we find the components to be b cos
c
h0.050 sg 51° = 11 m / s
= v sin 51° = c m / s h
340
0.050 sin
b sg 51° = 13 m / s v x = v cos 51° = 340 m / s 2
vy 4. 2 REAS...
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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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