Unit 4 Class Notes Accelerated Physics Projectile Motion Days 1 thru 3

Unit 4 Class Notes

Accelerated Physics

Projectile Motion

Days 1 thru 3

Day #1Free-Falling Object Review

In-Class Practice

In-Class Practice

In-Class Practice

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In-Class Practice

In-Class Practice

Day #2Horizontal Projectiles

From Reading Assignment handout

Horizontal Projectiles

• The first example on the previous slide is often called a horizontal projectile. This is the first example we will look at.– The object is given an initial horizontal velocity.

Thus, velocity is purely horizontal (no y component)

– Our initial velocity in the y direction viy=0.

– Next, analyze the motion in the horizontal and vertical direction separately.

Figure #1Figure #2

Figure #3

From Reading Assignment handout

Using figures from previous slide• Figure 1 – Free fall object. The object’s position after every

second of falling is shown. The object is accelerating because it travels a longer distance in each successive time interval.

• Figure 2 – Horizontally traveling object at a constant velocity. Acceleration equals zero so each time interval shows the same distance traveled.

• Figure 3 – A combination of both these motions. When a projectile falls, it accelerates downward so its y direction matches figure 1. However, since nothing (including air) slows the object down as it moves horizontally, its x direction matches figure 2.

Since projectile motion is a combination…• We will analyze the motion separately. The only shared variable between the two motions is TIME

(since they both occur in the same amount of time)

• Solve for time in vertical (y) side, then plug in to horizontal side.

Horizontal (x) Vertical (y)ax =0vix =vfx = constant = vx

∆x = vx∆t(distance = velocity x time)

ay=-9.8 m/s2

∆y = 1/2ay(∆t)2 +viy∆tVfy

2 = viy2 + 2ay∆y

A = (vfy – viy) /∆t∆y = ½∆t(viy +vfy)

Horizontal (x) Vertical (y)

x vxt

x (15 ms )t

x (15 ms )(2.474sec)

37.115m

y 12 ay (t)

2 v1yt

( 30m) 12 ( 9.8 m

s2 )(t)2 (0)t

solve for t 2.474sec

From Reading Assignment handout

•There are so few equations in the x direction because a=0 so velocity is constant.•In the y direction we have the 4 familiar acceleration formulas.

EXAMPLE 1: A ball is rolled off a flat roof hat is 30 m above the ground. If the ball’s initial speed is 15 m/s, find the time needed to strike the ground as well as the distance that it lands away from the building.

Horizontal (x) Vertical (y)

x vxt

9ft = vxt

y 12 ay (t)

2 v1yt

4ft 12 ( 32.2 m

s2 )t 2 (0)t

Solve for t = .498sec

vx (9ft)/(.498sec)

=18.072 fts

From Reading Assignment handout

Horizontal (x) Vertical (y)

x vxt

x (200 ms 150 m

s )(20s)

7,000 ft

y 12 ay (t)

2 v1yt

y 12 ( 9.8 m

s2 )(20s)2 (0)(20s)

Solve for y 1,960m

Therefore, the altitude is 1,960 mFrom Reading Assignment handout

In-Class Practice

In-Class Practice

In-Class Practice

Day #3COMPETITION LAB!!!! (Ball Rolling Off Table)

Day #4Tracking and Impact Velocity

Side-ways Toss (Gravity Turned Off)

Each position corresponds to 1 sec later than the previous position

(starting with the red dot)

Horizontal Projectiles

Side-ways Toss (Gravity Turned Off)

Free-Fall

Horizontal Projectile

Side-ways Toss (Gravity Turned Off)

Free-FallVelocities

Impact Velocity

Impact Velocity

Vfx

VfyVimpact

From Reading Assignment handout

From Reading Assignment handout

Horizontal (x) Vertical (y)

x vxt

x (4 ms )(2) 8m

y 12 ay (t)

2 v1yt

y 12 ( 9.8 m

s2 )(2)2 19.6m

tav yiy fyv

6.19)2)(8.9(0v 2fy sm

sm

19.6 m/s

4 m/s

V2

V2 = 20 m/s [78.46o BH]

From Reading Assignment handout

In-Class Practice

In-Class Practice

In-Class Practice

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Day #5Angled Projectiles on

Flat Ground

What we’ve done so far

What we still need to do

Projectiles at Angles

Side-ways Toss (Gravity Turned Off)

Throw-up, then Free-Fall

So, what are the Equations we use for this “NEW” type of

projectile????

Vertical Horizontal

tv

vv

a

x

fxix

x

vconstant

0

x

)t(

2

)(y

t

8.9

21

22

221

2

fyiy

iyfy

iy

iyfy

sm

vvy

yavv

tvta

avv

a

There are NO NEW EQUATIONS

vi

vi

vix

viy = visin

= vicos

So…..How are “angled” projectiles different than “horizontal” projectiles?

Viy = 0 (for horizontal projectiles)

Viy = something, + or -

(for horizontal projectiles)

viWhat are some important things to remember?

The velocity at every moment in time is the resultant of the two velocity components

Vix = Vfx = Vx = 0 (since ax = 0)

When rising, Vy is positive (when falling = Vy is negative)

Vy, top = 0

If launching and landing occurs at the same height (on level ground) then the launching/landing speeds & angles are the same

In-Class Practice

In-Class Practice

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Day #6MORE Angled Projectiles …

Sometimes NOT on Flat Ground

In-Class Practice

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In-Class Practice

Day #7Field Goal Style Problems

In-Class Practice

In-Class Practice

In-Class Practice