Question: If a ball is released from rest and another is shot horizontally from the same height, which hits the ground first?
Answer: C) Both hit at the same time.
Explanation: Vertical motion is independent of horizontal motion. Video explanation at: https://www.youtube.com/watch?v=zMF4CD7i3hg (at 1:00)
Gravitational Field
Direction: Downward.
Magnitude: g=9.8s2m near Earth’s surface.
Decreases with distance from Earth’s center.
g is four times smaller if twice as far from the center.
Projectile Motion Details
Basic Idea: Free fall with horizontal motion.
Assumptions:
Negligible air resistance.
Not too fast or light.
Uniform gravitational field, g.
Near Earth's surface.
Small trajectories compared to Earth.
g≈9.8s2m
Projectile Motion Tools
Acceleration components:
a=g
ax=0
ay=−g
g=9.8s2m
2D kinematic equations with constant acceleration during free fall.
Trajectory
The trajectory of projectile motion is a downward parabola.
Problem Solving Steps
Diagram the problem:
Include the trajectory.
Identify time intervals of constant acceleration.
Label moments where you know information.
Label moments where you want to know information.
List Knowns and Unknowns: At each significant moment.
Decide which kinematic equations will convert knowns into desired unknowns.
Problem Solving Tips
Time relates horizontal and vertical motion (it's the only common variable).
Unstated Information:
Max height implies vy=0
Known acceleration in projectile motion.
Be careful with timing:
Differentiate between just before vs just after landing.
Velocity after hitting the ground is usually irrelevant.
Example Problem: Fire Fighting
Scenario: A firefighter needs to spray water into a 1 m tall window 12 m above the ground. The nozzle is 1 m above the ground and 20 m from the building. If the water exits the nozzle at 35.0° above the horizontal at a speed of 30 m/s, will it enter the window?
Solution: No, the water hits at 11.8 m above the ground, which is 0.2 m below the window.
Class Problem: Is it a Homerun?
Scenario: A baseball is hit at a 40° angle from a height of 1 m above the ground. There is a 4 m tall wall 100 m away. Ignore air resistance.
Question: How fast was the baseball moving right after it was hit to barely clear the wall?
Answer: 32.1 m/s (approximately 72 mph).
Horizontal Range
Requirements:
Same start and end height (level ground).
Free fall.
Formula: R=gv<em>i2sin2θ</em>i
R = Range
vi = Initial velocity
θi = Launch angle from horizontal
g = Gravitational field magnitude
Maximum Height
Basic Idea: How high the projectile reaches above launch height.
Requirements: Free fall and level ground.
Formula: y<em>max=2gv</em>i2sin2θi
ymax = Max height
vi = Initial velocity
θi = Launch angle from horizontal
g = Gravitational field magnitude
Clicker Question 11
Question: On Earth, the longest recorded golf drive was 471 m. About how far would it have gone on the moon, where gravity is 6 times weaker?
Answer: C) 471(6) m
Example Problem: The Long Jump
Scenario: A daredevil barely jumps over 24 trucks (each 3 m wide) off a 30° ramp.
Question: How fast was he going when he left the ramp?
Answer: 28.5 m/s or 63.8 mph.
Class Problem: Cannonball!
Scenario: A cannon is fired at an angle of 60° across level ground. Neglecting air resistance, the cannonball leaves the barrel at 90 m/s.
Questions:
How high does the cannonball reach?
How far does the cannonball go before hitting the ground?
Answers:
Height: 310 m
Distance: 716 m
Circular Motion
Basic Concepts
Easier to use directions relative to the circle's center.
Definitions:
Centripetal (c): Toward the center.
Tangential (t): Counterclockwise along the circle.
Use c & t for components like x and y.
Circular Motion: Constant Speed
Change in velocity is toward the center over short times.
Acceleration is toward the center = “Centripetal” component, ac.
Formula: ac=rv2
ac = Centripetal acceleration
v = Speed
r = Radius of the circle
Circular Motion: Changing Speed
Turning requires centripetal acceleration, ac (how quickly velocity is turning).