Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter4: Two-and-three Dimensional Motion
Section: Chapter Questions
Problem 47PQ
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Topic Video
Question
Only practice problem 3.3
Second photo is background info to help solve problem
Transcribed Image Text:Chapter 3: Practices Problems
Name:
G#
Page 66- Practice Problem 3.1
Suppose you reverse the car's motion, so that it retraces its
path in the opposite direction in the same time. Find the
components of the average velocity of the car and the
magnitude and direction of the average velocity. Answer:
-6.0m/s, -8.0m/s, -10.0m/s, 233⁰
0
Page 69- Practice Problem 3.2
Suppose that as the car continue to move, it has velocity
components v= 3.5 m/s and v, 1.0 m/s at time t,= 3.0s.
What are the magnitude and direction of the average
acceleration between t₂=2.0 s and t₂= 3.0s? Answer: 4.7m/s,
-58⁰ from +x.
→ Page 73- Practice Problem 3.3
If air resistance is ignored, what initial speed is required
for a range of 20m? Answer: 36.1 m/s.
Page 74- Practice problem 3.4
If the ball could continue to travel below its original level
(through an appropriately shaped hole in the ground), then
negative values of y corresponding to times greater than
6.04s would be possible. Compare the ball's position and
velocity 8.00s after the start of its flight. Answers: x=178m.
y=-76.8m, v, 22.2m/s, v,=-48.8 m/s.
Page 75- Practice Problem 3.5
Show the range is the same when the launch angle is 30° as
when it is 60° and that the range for any launch angle is
the same as for the complementary angle (90°-
Answer: sinsin(180-), so sin2000 sin2(90°-
Transcribed Image Text:74
CHAPTER 3 Motion in a Plane
EXAMPLE 3.3 Paintball gun
Now we are going to look at a classic example of projectile motion: a bullet fired from a gun. As shown in
Figure 3.13, a paintball is fired horizontally at a speed of 75.0 m/s from a point 1.50 m above the ground. The
ball misses its target and hits the ground some distance away. (a) For how many seconds is the ball in the air?
(b) Find the maximum horizontal displacement (which we'll call the range of the ball). Ignore air resistance.
SOLUTION
0
SET UP We choose to place the origin of the coordinate system at
ground level, directly below the end of the gun barrel, as shown in
Figure 3.14. (This choice of position for the origin avoids having to
deal with negative values of y, a modest convenience.) Then xo
and yo= 1.50 m. The gun is fired horizontally, so vor= 75.0 m/s and
Voy = 0. The final position of the ball, at ground level, is y = 0.
SOLVE Part (a): We're asked to find the total time the paintball is in
the air. This is equal to the time it would take the paintball to fall vertically
from its initial height to the ground. In each case, the vertical position
is given as a function of time by Equation 3.11, y = yo voyt - gt².
In this problem, yo = 1.50 m, y = 0, and voy 0, so that equation be-
comes simply
0 = 30 - /gt².
A FIGURE 3.13
Dikk
and mor
We need to find the time t. Solving fort and substituting numeric
values, we obtain
1 =
210
8
2(1.50 m)
9.80 m/s²
Video Tutor Se
= 0.533 s.
Part (b): Now that we know the time t of the ball's flight through the a
we can find the range-that is, the horizontal distance x it travels during
time t. We use Equation 3.9: x = xo + Vaxt. Setting xo = 0, we find
x = xo + Voxt = 0 + (75.0 m/s) (0.553 s) = 41.5 m.
REFLECT Actual ranges of paintballs are less than this, typically abou
30 m. The difference is due primarily to air resistance, which decrease
the horizontal component of velocity.
Yo = 1.50 m
Practice Problem: If air resistance is ignored, what initial speed is
required for a range of 20 m? Answer: 36.1 m/s.
Vox = 75.0 m/s
Y=0
0¹x0=0
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