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


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.


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.


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.



• STEP 1: Coordinates! (and signs!)

• +x horizontally to the right

• +y vertically upwards

• acceleration of gravity = -9.8 m/s2



• 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



• STEP 3: LOOK FOR 3 THINGS from!

• v = v0 + at

• Dx = ½ (v0 + v)*t

• Dx = v0*t + ½ at2

• v2 = v02 +2aDx



• In x direction, ax = 0!

• vx = v0x

• Dx = v0x*t = v cosq * t



• 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



• 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



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



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



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


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)



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



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



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.



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?



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.


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?


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?


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?


Tranquilizing a falling monkey

• Where should the zookeeper aim?

• Follow Example 3.10.


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.

03 Lecture Outline Part II

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