1. Periods 2 and 3 Take notes on the following in your Physics Journals
HW will be handed out, due on FRIDAY
Look at Notes PPT and Practice Probs online
TEST on Projectiles is TUESDAY 11.19

2. REVIEW: Do both of these equations still apply?
NO!!! Since this is launched at an angle viy ≠ 0 m/s, so you have to use the full ∆ 𝒙 𝒚 = 𝒗 𝒊𝒚 ∆𝒕+ 𝟏 𝟐 𝒂 𝒚 ∆ 𝒕 𝟐 to find vertical displacement from launch point.
∆ 𝑥 𝑦 = 1 2 𝑎 𝑦 ∆ 𝑡 2 ∆ 𝑥 𝑥 = 𝑣 𝑥 ∆𝑡
YES!!! Since there are no horizontal forces, ax = 0 m/s, then you can use ∆ 𝑥 𝑥 = 𝑣 𝑥 ∆𝑡 to find the horizontal displacement.

3. ay = -10 m/s²
Take 4 min to solve in Journals!
Time vx
∆xx vy
∆xy 1 2 3 4 5 6
Let’s practice using these equations. The diagram below shows the trajectory for a projectile launched NON-horizontally. The initial vertical and horizontal components are given. Complete the chart given for the scenario.
Note: this is vertical displacement from starting point (do not confuse with height off the ground!)

4. ay = -10 m/s²
Time vx
(m/s)
∆xx
(m) vy
∆xy 8 20 1 10 15 2 16 3 24
-10 4 32
-20 5 40
-30
-25 6 48
-40
-60
Let’s practice using these equations. The diagram below shows the trajectory for a projectile launched NON-horizontally. The initial vertical and horizontal components are given. Complete the chart given for the scenario.

5. Vector Addition By Components
11/13/2013

6. 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.

7. 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 𝐶𝑜𝑠𝜃= 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑒𝑢𝑠𝑒 𝑇𝑎𝑛𝜃= 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡

8. Adding perpendicular vectors using Pythagorean Theorem:
Place vectors tail to head and use

9. Solve for the resultant using the Pythagorean Theorem
All resultants will be the same
All resultants will be the same: 5m

10. 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
Using the Pythagorean Theorum, this is simplifies to a 12 by 5 right triangle….
square root of (12² + 5²) = 13 meters

11. What about the directions?
Use SOH CAH TOA to determine the direction (angle) of the resultant.
sin Θ = 𝟓 𝟏𝟑
Θ = 𝐬𝐢𝐧 −𝟏 ( 𝟓 𝟏𝟑 )
Θ = 22.6 ° Θ
Use SOH CAH TOA to determine the direction (angle) of the resultant.

12. What happens if you are ONLY given the resultant’s magnitude and direction?
-Find the component vectors in the x- and y- direction.
-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.
A component is the effect of a vector in a given x- or y- direction. A component can be thought of as the projection of a vector onto the nearest x- or y-axis. 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.

15. 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.
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 m, 14.5°South of East

16. 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)

17. Homework…. Due tomorrow for a grade.