Periods 2 and 3
Take notes on the following in your
Physics Journals
11.13.2013
• HW will be handed out, due on FRIDAY
• Look at Notes PPT and Practice Probs online
• TEST on Projectiles is TUESDAY 11.19
REVIEW: Do both of these equations still apply?
∆𝑥𝑦 =
NO!!! Since this is launched at an angle
1
𝑎𝑦 ∆𝑡 2 viy ≠ 0 m/s, so you have to use the full
2
𝟏
∆𝒙𝒚 = 𝒗𝒊𝒚 ∆𝒕 + 𝒂𝒚 ∆𝒕𝟐 to find vertical
𝟐
displacement from launch point.
∆𝑥𝑥 = 𝑣𝑥 ∆𝑡
YES!!! Since there are no horizontal forces, ax = 0 m/s, then
you can use ∆𝑥𝑥 = 𝑣𝑥 ∆𝑡 to find the horizontal displacement.
ay = -10 m/s²
Let’s practice using these equations.
Take 4 min
to solve
The diagram below shows
thein
Journals!
trajectory for a projectile launched
Time
0
1
2
3
4
5
6
vx
∆xx
vy
∆xy
NON-horizontally. The initial vertical
Note: this isare
and horizontal components
vertical
displacement
given. Complete the chart
givenfrom
for
starting point (do
the scenario. not confuse with
height off the
ground!)
ay = -10 m/s²
Let’s practice using these equations.
The diagram below shows the
Time
v for
∆x a projectile
v
∆x launched
trajectory
(m/s)
NON-horizontally.
The
initial
vertical
0
8
0
20
0
1
8
8
10
15
and
horizontal
components
are
2
8
16
0
20
given.
Complete
the
chart
given
for
3
8
24
-10
15
the
4
8
32 scenario.
-20
0
x
5
6
x
(m/s)
(m)
8
8
40
48
y
y
(m)
-30
-40
-25
-60
Vector Addition By Components
11/13/2013
Non-Horizontally Launched Projectiles
In order to analyze the motion of a
non-horizontally launched projectile, you MUST
be able to look at the components of that
motion critically.
In order to do so we need to utilize some more
complex, but familiar math.
General Notes about Math for Vector Addition
Remember SOH-CAH-TOA!
Trig Function are mathematical functions that
relate the length of the sides of a right triangle to
the angles of the triangle. The meaning of the
three functions can easily be remembered using
the mnemonic:
SOH CAH TOA
𝑆𝑖𝑛𝜃 =
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑒𝑢𝑠𝑒
𝑇𝑎𝑛𝜃 =
𝐶𝑜𝑠𝜃 =
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑒𝑢𝑠𝑒
Adding perpendicular vectors using
Pythagorean Theorem:
• Place vectors tail to head and use
Solve for the resultant using the
Pythagorean Theorem
All resultants will be the same: 5m
Okay, so those were pretty easy! Now try this…
WTH?!
Using the Pythagorean
Theorem, this
simplifies to a 12 by 5
right triangle…
*Copy this and try to
calculate the answer in
your journals
square root of (12² + 5²) = 13 meters
What about the directions?
Use SOH CAH TOA to determine the
direction (angle) of the resultant.
Θ
sin Θ =
𝟓
𝟏𝟑
𝟓
𝟏𝟑
Θ = 𝐬𝐢𝐧−𝟏 (
Θ = 22.6 °
)
What happens if you are ONLY given the resultant’s
magnitude and direction?
-Find the component
vectors in the x- and ydirection.
-SOH CAH TOA allows a
student to determine a
component from the
magnitude and direction
of a vector.
-Determine the
components of the
following vectors in your
journals.
PRACTICE: Consider the following vector diagrams for the
displacement of a hiker. For any angled vector, use SOH CAH TOA to
determine the components. Then sketch the resultant and
determine the magnitude and direction of the resultant.
Solve in Journal!
The resultant is the hypotenuse of a 18.9 m and 4.9 m right triangle.
19.5 m, 14.5°South of East
Solving word problems by drawing
vectors.
Jennifer’s dog got out this morning and she is
searching for him. She walks 9 km N and 17 km
W. What is the distance that she traveled? What
was the magnitude and direction of her
displacement?
19.2 m, 62° “West of North” (i.e. 28 ° N of W)
Homework….
Due tomorrow for a grade.