Unit 4 Class Notes Accelerated Physics Projectile Motion Days 1 thru 3.
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Transcript of 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
In-Class Practice
In-Class Practice
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
In-Class Practice
In-Class Practice
In-Class Practice
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
In-Class Practice
In-Class Practice
In-Class Practice
Day #6MORE Angled Projectiles …
Sometimes NOT on Flat Ground
In-Class Practice
In-Class Practice
In-Class Practice
Day #7Field Goal Style Problems
In-Class Practice
In-Class Practice
In-Class Practice