1. Final Velocity Free Fall 2D Motion
Final Exam Review
Final Velocity
Free Fall
2D Motion

2. Final Velocity When solving for final velocity use:
Formula: vf2 = vi2 + 2aΔx

3. Example
A person pushing a stroller starts from rest, uniformly accelerating at a rate of m/s2. What is the velocity of the stroller after it has traveled 4.75m?

4. Example
A person pushing a stroller starts from rest, uniformly accelerating at a rate of m/s2. What is the velocity of the stroller after it has traveled 4.75m?
Givens a = m/s vi = 0 m/s
Δx = 4.75 m

5. Example
A person pushing a stroller starts from rest, uniformly accelerating at a rate of m/s2. What is the velocity of the stroller after it has traveled 4.75m?
Givens a = m/s vi = 0 m/s
Δx = 4.75 m
vf2 = vi2 + 2aΔx
vf2 = (0m/s)2 + 2(0.500 m/s2)(4.75m)
vf2 = → vf = 2.18 m/s

6. Practice Practice E- Pg. 58 #3-4 ONLY
Hint: For questions that require solving for more than vf, see which formula we have already used that will help.

7. Free Fall In free fall problems, a = 9.8m/s2 every time.
Formulas: vf = vi + at
vf2 = vi2 + 2aΔy
Motion is along the y axis. y x

8. Example
A pebble is dropped down a well and hits the water 1.5s later. Using the equations for motion with constant acceleration, determine the distance from the edge of the well to the water’s surface.

9. Example
A pebble is dropped down a well and hits the water 1.5s later. Using the equations for motion with constant acceleration, determine the distance from the edge of the well to the water’s surface.
Givens t = 1.5s a = 9.8 m/s vi = 0m/s
Unknowns Δy = ? vf = ? (need to find vf before Δy)

10. Example
A pebble is dropped down a well and hits the water 1.5s later. Using the equations for motion with constant acceleration, determine the distance from the edge of the well to the water’s surface.
Givens t = 1.5s a = 9.8 m/s vi = 0m/s
Unknowns Δy = ? vf = ? (need to find vf before Δy)
Step 1: vf = vi + at → vf = (0m/s) + (9.8m/s2)(1.5s) = 14.7m/s

11. Example
A pebble is dropped down a well and hits the water 1.5s later. Using the equations for motion with constant acceleration, determine the distance from the edge of the well to the water’s surface.
Givens t = 1.5s a = 9.8 m/s vi = 0m/s
Unknowns Δy = ? vf = ? (need to find vf before Δy)
Step 1: vf = vi + at → vf = (0m/s) + (9.8m/s2)(1.5s) = 14.7m/s
Step 2: vf2 = vi2 + 2aΔy → (14.7)2 = (0m/s)2 + 2(9.8m/s2)Δy
Δy = 11 m

12. Practice
Practice F- Pg 64
FIRST TWO ONLY

13. 2D Motion Key Point: Draw a diagram
Key Point: Only consider one direction at a time (either “x” or “y”)
Key Point: Step one will almost always be to solve for time.

14. Example
The Royal Gorge Bridge in Colorado rises 321m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock’s horizontal displacement is 45.0m. Find the speed at which the rock was kicked.

15. Example Givens x direction y direction Δx = 45m Δy = -321m
The Royal Gorge Bridge in Colorado rises 321m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock’s horizontal displacement is 45.0m. Find the speed at which the rock was kicked.
Givens
x direction y direction
Δx = 45m Δy = -321m
vi=? a = -9.8 m/s2
t = ? Time will always be the same t = ?

16. Example- Page 98 (Diagram)
Givens
x direction y direction
Δx = 45m Δy = -321m
vi=? a = -9.8 m/s2
t = ? Time will always be the same t = ?
Step 1: Solve for time.
Δy = vit + ½at2 → -321m = (0m/s)(t) + ½(-9.8m/s2)t2 → t2 = 65.5
t = 8.09 s

17. Example- Page 98 (Diagram)
Givens
x direction y direction
Δx = 45m Δy = -321m
vi=? a = -9.8 m/s2
t = 8.09 s t = 8.09 s
Step 1: Solve for time.
Δy = vit + ½at2 → -321m = (0m/s)(t) + ½(-9.8m/s2)t2 → t2 = 65.5
t = 8.09 s
Step 2: Solve for vi in the x-direction.
Δx = vit → 45m = vi(8.09s)
vi = 5.56 m/s

18. Practice
Practice D- Pg 99
FIRST TWO