Transcript Chapter 7 Gravitation - REDIRECT TO NEW SITE

In this chapter you will:
 Use Newton’s laws and your knowledge of
vectors to analyze motion in two
dimensions.
Solve problems dealing with projectile
and circular motion.
Solve relative velocity.
CHAPTER 6 SECTIONS
 Section 6.1: Projectile Motion
 Section 6.2: Circular Motion
 Section 6.3: Relative Velocity
SECTION 6.1 PROJECTILE MOTION
 Objectives
 Recognize that the vertical and horizontal
motions of a projectile are independent.
 Relate the height, time in the air, and initial
vertical velocity of a projectile using its vertical
motion, and then determine the range using the
horizontal motion.
 Explain how the trajectory of a projectile
depends upon the frame of reference from
which it is observed.
INTRO
 Parabola – set of all points equidistant from a fixed line
called the directix, and a fixed point not on the line called
the focus. Will be a U shaped graph.
 Projectile – an object shot through the air, such as a
football, that has Independent Vertical and Horizontal
motions and after receiving an initial thrust travels through
the air only under the force of gravity. From Old Book,
motion of objects given an initial velocity that then
move only under the force of gravity.
 Trajectory – the path of a projectile through space.
INDEPENDENCE OF MOTION IN
TWO DIMENSIONS
 Example of dropping a softball and launching one horizontally at 2 m/s.
In both cases the horizontal acceleration is ZERO. (Dropped ball does
not move horizontally and launched ball had a constant velocity thus
no acceleration. (See Figure 6-1)
 Also in this example you see that the dropped and launched balls have
the same vertical motion. Both balls are accelerated downward by the
force of gravity. Both balls would hit the ground at the same time.
Similar to the boat taking the same time to get across the river if there
were no flow downstream and if there was a current downstream.

 The horizontal motion of the thrown ball does not affect its vertical
motion at all. The horizontal and vertical components are
Independent of each other.
INDEPENDENCE OF MOTION IN
TWO DIMENSIONS
 The combination of a Constant Horizontal Velocity
and Uniform Vertical Acceleration (Gravity) produces
a Trajectory that has a Parabolic Shape.
 Since Horizontal and Vertical parts are Independent
of each other if you find the time of one the other is
the same, similar to the boat across the river
problems.
 The shape of the trajectory and the horizontal motion
depend on the viewpoint or frame of reference of the
observer, But the Vertical Motion does not.
INDEPENDENCE OF MOTION IN
TWO DIMENSIONS
 x = vxt
where x is the horizontal displacement, vx is
the initial horizontal velocity, and t is the time
 vxf = vi
where vxf is the final horizontal velocity and vi
is the initial velocity
 y = v yt + ½ gt2
where y is the vertical displacement, v y
is the initial vertical velocity, t is time, and g is gravity
 v yf = v y + gt where v yf is the final vertical velocity and v y is
initial vertical velocity
 Do Practice Problems p. 150 # 1-3
PROJECTILES LAUNCHED AT AN
ANGLE
 When a projectile is launched at an angle the initial velocity has a
vertical and horizontal component.
 Max Height – the height of the projectile when the vertical velocity is
zero. The max height occurs when the time is HALF of the entire flight
time (Example if time is 5 seconds then the time for the max height is
2.5 seconds).
 Range – denoted by R; the horizontal distance between the launch
point of the projectile and where it returns to launch height; or the
horizontal distance from the point of bounce until the projectile
returns to the surface height.
 Range is the horizontal distance traveled during the entire flight
time.
PROJECTILES LAUNCHED AT AN
ANGLE
 vx = vi cos 
 v y = vi sin 
or
or
Ax = Ai cos 
Ay = Ai sin 
PROJECTILES LAUNCHED AT AN
ANGLE
 Do Example 1 p. 151
 vx = vi cos  = 4.5(cos 66) = 4.5(.407) = 1.83 m/s
 v y = vi sin  = 4.5(sin 66) = 4.5(.914) = 4.11 m/s
 A) Time
y = v yt + ½ gt2
0 = v yt + ½ gt2
-4.11t = ½ (-9.8)t2
t
t
-4.11 = -4.9t
-4.11 / -4.9 = t
.839 s = t
C) On next slide
b) Max Height
Since the trajectory is symmetric
the max height occurred at .4165 s
So y = v yt + ½ gt2
y = 4.11(.4195) + ½ (-9.8)(.4195)2
y = 1.724 + (-4.9)(.176)
y = 1.724 - .862
y = .862 m
PROJECTILES LAUNCHED AT AN
ANGLE
 c) Range is the horizontal distance traveled during the entire
flight time.
So R = vxt
R = 1.83(.839)
R = 1.535 m
 Do Practice Problems p. 152 # 4-6
TRAJECTORIES DEPEND UPON
THE VIEWER
 Remember that the force due to air resistance exists and it can
be important but for now we are ignoring it.

 Do 6.1 Section Review p. 152 # 7-11