Download - 03 Lecture Outline Part II
Copyright © 2012 Pearson Education Inc.
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by Wayne Anderson
Chapter 3
Motion in Two or
Three Dimensions
Part II
Copyright © 2012 Pearson Education Inc.
Projectile Motion
If object is launched at initial angle θ0 with the horizontal,
analysis is similar except that the initial velocity has a
vertical component.
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
Example: A kicked football.
A football is kicked at an angle θ0 = 37.0° with a velocity
of 20.0 m/s, as shown.
It travels through the air (assume no resistance) and
lands at the ground.
Copyright © 2012 Pearson Education Inc.
Example: A kicked football.
Calculate
a) the maximum height,
b) the time of travel before the football hits the ground,
c) how far away it hits the ground.
d) the velocity vector at the maximum height,
e) the acceleration vector at maximum height.
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
• STEP 1: Coordinates! (and signs!)
• +x horizontally to the right
• +y vertically upwards
• acceleration of gravity = -9.8 m/s2
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
• STEP 2: COMPONENTS! (and signs!)
• +vx0 horizontally to the right = v0 cos q
• +vy0 vertically upwards = v0 sinq
• ay acceleration of gravity = -9.8 m/s2 only in y
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
• STEP 3: LOOK FOR 3 THINGS from!
• v = v0 + at
• Dx = ½ (v0 + v)*t
• Dx = v0*t + ½ at2
• v2 = v02 +2aDx
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
• In x direction, ax = 0!
• vx = v0x
• Dx = v0x*t = v cosq * t
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
• In y direction, ay = -9.8!
• vy = v0y – 9.8t = v0 sin q - gt
• Dy = ½ (v0y + vy)*t
• Dy = v0y*t – 4.9t2 = (v0 sin q) t - ½ gt2
• vy2 = v0y
2 – 19.6Dy = (v0 sin q)2 - 2gDy
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
• In this problem, x0 = y0 = 0
• the equations describing
projectile motion in this case are
shown at the right.
(1)
(2)
(3)
(4)gtvv
vv
gttvy
tvx
y
x
00
00
2
00
00
sin
cos
2
1)sin(
)cos(
q
q
q
q
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
Solve it!
• v0x = 16 m/s
• v0y = 12 m/s
• Max height = 7.3 m
• Max distance in x (“range”) = 39 m
• Time = 2.5 seconds
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
Details of calculating Max height
Conditions at Max height: vy = 0 m/s
From Eq(4), we get time: t = v0sinθ0/g
By replacing t in Eq(2), we get:
Max height = (v02sin2θ0)/2g = 7.3 m
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
Calculating the range (condition at the range: y = 0)
From Eq(2): 0 = (v0sinθ0)t – (1/2)gt2
=> t(v0sinθ0– (1/2)gt) = 0 = either t => 0 or t = (2 v0sinθ0)/g
By replacing t in Eq(1), we get:
Range = 2cosθ0sinθ0v02/g = (v0
2/g)sin(2 θ0) = 39 m
Copyright © 2012 Pearson Education Inc.
Example: Level horizontal range.
The horizontal range is
defined as the horizontal
distance the projectile travels
before returning to its
original height (which is
typically the ground); that is,
y(final) = y0.
(a) Derive a formula for the
horizontal range R of a
projectile in terms of its
initial speed v0 and angle θ0.
(solved in the previous slide!)
Range = (v02/g)sin(2 θ0)
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
Example: Level horizontal range.(b) Suppose one of Napoleon’s
cannons had a muzzle speed, v0,
of 60.0 m/s.
At what angle should it have been
aimed (ignore air resistance) to
strike a target 320 m away?
R = (v02/g)sin(2 θ0)
sin(2 θ0) = gR/v02
sin(2 θ0) = 9.8*320/602 = 0.87
=> 2 θ0 = 60o => θ0 = 30o
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
Example: Maximum range.
(c) At what angle should it have
been aimed (ignore air
resistance) to strike a target at
the maximum range away?
Max Range = (v02/g)sinmax(2 θ0)
Max range obtained for θ0˛= 45o
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
Projectile motion is parabolic:
Taking the equations for x and y as a function
of time, and combining them to eliminate t, we
find y as a function of x:
This is the equation for a parabola.
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
Suppose the football was punted and left the
punter’s foot at a height of 1.00 m above the
ground. How far did the football travel before
hitting the ground?
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
The wrong strategy!
A boy on a small hill aims his water-balloon slingshot
horizontally, straight at a second boy hanging from a
tree branch a distance d away. At the instant the water
balloon is released, the second boy lets go and falls
from the tree, hoping to avoid being hit. Show that he
made the wrong move. (He hadn’t studied physics
yet.) Ignore air resistance.
Copyright © 2012 Pearson Education Inc.
Rescue helicopter drops supplies.
A rescue helicopter wants to drop a package of supplies to isolated
mountain climbers on a rocky ridge 200 m below.
If the helicopter is traveling horizontally with a speed of 70 m/s (250
km/h), (a) how far in advance of the recipients (horizontal distance)
must the package be dropped?
Copyright © 2012 Pearson Education Inc.
Rescue helicopter drops supplies.
(b) Suppose, instead, that the helicopter releases the package a
horizontal distance of 400 m in advance of the mountain climbers.
What vertical velocity should the package be given (up or down) so
that it arrives precisely at the climbers’ position?
(c) With what speed does the package land in the latter case?
Copyright © 2012 Pearson Education Inc.
Height and range of a projectile
• A baseball is batted at an angle.
• Vo = 37.0 m/s at 53.1 degrees; @ 2 seconds where is it?
Copyright © 2012 Pearson Education Inc.
Tranquilizing a falling monkey
• Where should the zookeeper aim?
• Follow Example 3.10.
Copyright © 2012 Pearson Education Inc.
The effects of air resistance—Figure 3.20
• Calculations become
more complicated.
• Acceleration is not
constant.
• Effects can be very
large.
• Maximum height and
range decrease.
• Trajectory is no longer
a parabola.