FULL REVIEW
Full Review: Projectile Motion — Foundational
Review the essential ideas, relationships, and problem-solving tools for projectile motion.
TIME
45–60 minutes
BEST FOR
A complete topic review
FINISH WITH
A readiness check
After this full review, you'll be able to...
recall the essential ideas, apply them to representative problems, and determine what to study next.
Choose how you want to review
Course Alignment
This bundle is designed to complement the chapter listed below. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.
TEXTBOOK: Independent Physics Sensei Unit Review
CHAPTER: MEC-U15
TOPIC: MEC-U15 — Projectile Motion
COURSE LEVEL: Foundational introductory physics
BEST USED
✓ After reading the chapter
✓ Before starting homework
✓ Before a quiz or exam
Physics Sensei is an independent educational resource for college physics review and practice.
Your Review Plan
Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.
6 Stages • Approximately 45–60 minutes.
Warm-Up Check
Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.
ACTIVITY 1
Recall Activity
Use the motion diagram, not a formula, to identify which parts of the motion change.
A particle moves from r1 = <2, −1, 3> m to r2 = <8, 4, −1> m. Find the displacement vector and its magnitude.
Reveal Answers
Δr = <6, 5, −4> m; |Δr| = √77 m = 8.77 m.
Why it works: Displacement is final position minus initial position, component by component: <8 − 2, 4 − (−1), −1 − 3>. Its magnitude is √(62 + 52 + (−4)2).
ACTIVITY 2
Recall Activity
Think about the top of the path as a turning point in only one direction.
A projectile has velocity components vx = 12.0 m/s and vy = 18.0 m/s at launch. Neglect air resistance. Find its velocity components, speed, and direction 1.50 s later. Use g = 9.80 m/s2.
Reveal Answers
vx = 12.0 m/s; vy = 3.30 m/s; speed = 12.4 m/s; direction = 15.4° above +x.
Why it works: Horizontal acceleration is zero, so vx stays 12.0 m/s. Vertically, vy = 18.0 − (9.80)(1.50) = 3.30 m/s. The speed is √(12.02 + 3.302) = 12.4 m/s, and atan2(3.30, 12.0) = 15.4°.
ACTIVITY 3
Recall Activity
Compare two objects that have identical vertical starting conditions.
Two balls leave the same height at the same instant. One is dropped and one is launched horizontally. Which reaches the floor first if air resistance is negligible?
Reveal Answers
The acceleration has magnitude 9.00 m/s2 and points radially inward, toward the center.
Why it works: Gravity acts vertically throughout ideal projectile motion, so ax = 0 and ay = −g. Horizontal velocity stays constant while the vertical velocity changes uniformly.
Ready to strengthen your understanding?
You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?
Core Concepts
Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.
KEY CONCEPT 1
The Ideal Projectile Model
Projectile motion is two perpendicular motions happening at once. With air resistance neglected, gravity is the only acceleration: horizontal motion is uniform while vertical motion is uniformly accelerated downward. The two components are physically independent but occur during the same elapsed time.
r(t) = <x(t), y(t), z(t)>; Δr = r2 − r1; vavg = Δr/Δt; v(t) = dr/dt; a(t) = dv/dt; speed = √(vx2 + vy2 + vz2).
Example: If r(t) = <3t2, 4t, 2 − t2> m, then v(t) = <6t, 4, −2t> m/s and a(t) = <6, 0, −2> m/s2. At t = 2.00 s, v = <12, 4, −4> m/s and the speed is √176 = 13.3 m/s.
Sensei Note: A vector derivative is taken component by component. Never differentiate the magnitude of r and call that the velocity vector; doing so discards direction.
KEY CONCEPT 2
Trajectory, Time, Range, and Height
The curved path comes from combining constant horizontal motion with accelerated vertical motion. Time is the bridge between the axes. For a horizontal launch, the vertical drop determines how long the projectile is airborne; for an angled launch, the vertical component determines when it rises, reaches the top, and lands.
v0,x = v0 cos θ; v0,y = v0 sin θ; x = x0 + v0,xt; y = y0 + v0,yt − ½gt2; vy = v0,y − gt. For equal heights: T = 2v0,y/g, R = v02 sin(2θ)/g, H = v0,y2/(2g).
Example: A ball launched at 20.0 m/s and 35.0° from a height of 1.80 m has v0,x = 16.4 m/s and v0,y = 11.5 m/s. Solving 0 = 1.80 + 11.5t − 4.90t2 gives t = 2.49 s. The horizontal range is 40.8 m. The maximum height above the ground is 8.51 m.
Sensei Note: A longer horizontal distance does not mean a longer fall time when the two objects have the same vertical starting conditions.
KEY CONCEPT 3
Velocity Along the Path and Model Limits
Velocity is tangent to the trajectory. The horizontal component remains constant, while the vertical component changes continuously because of gravity. The projectile speeds up while descending because its downward velocity grows. Real projectiles deviate from this model when air drag, lift, wind, or large altitude changes matter.
arad = v2/r = ω2r (inward); atan = dv/dt; |a| = √(arad2 + atan2); vA/C = vA/B + vB/C.
Example: A ball launched at 18.0 m/s and 40.0° from 5.00 m above the ground has v0x = 13.8 m/s and v0y = 11.6 m/s. Solving the vertical equation gives a flight time of 2.77 s and a horizontal range of 38.2 m.
Sensei Note: Acceleration is not required to point in the direction of motion. For a projectile, velocity is tangent to the path while acceleration is vertical downward.
Ready to apply these ideas?
You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?
Guided Practice
Now it's time to apply what you've reviewed.
Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.
PRACTICE 1
Worked Example
Use component reasoning to compare the two motions.
A particle has r(t) = <3t2 − 2, 4t + 1, −t2 + 6t> m. Find r, v, a, and speed at t = 2.00 s.
Reveal Answers
r = <10, 9, 8> m; v = <12, 4, 2> m/s; a = <6, 0, −2> m/s2; speed = √164 = 12.8 m/s.
Why it works: Differentiate component by component: v(t) = <6t, 4, −2t + 6> and a(t) = <6, 0, −2>. Substitute t = 2.00 s. The speed is the magnitude of v, √(122 + 42 + 22).
PRACTICE 2
Guided Problem
Track which component changes before, at, and after the top of the trajectory.
A ball is launched from a 1.80 m platform at 20.0 m/s and 35.0° above horizontal. Neglect air resistance. Find the flight time, horizontal range, maximum height above the ground, and impact velocity.
Reveal Answers
Flight time = 2.49 s; range = 40.8 m; maximum height = 8.51 m; impact velocity = <16.4, −12.9> m/s, with speed 20.9 m/s at 38.3° below +x.
Why it works: Use v0,x = 16.4 m/s and v0,y = 11.5 m/s. The positive root of 0 = 1.80 + 11.5t − 4.90t2 is 2.49 s. Then x = v0,xt. The rise above launch is v0,y2/(2g). At impact, vy = v0,y − gt = −12.9 m/s.
PRACTICE 3
Independent Problem
Decide whether the ideal projectile model is appropriate.
For each case, state whether the ideal projectile model is a reasonable first approximation: a tossed classroom ball, a feather released outdoors, and a long-range golf drive in strong wind.
Reveal Answers
Flight time = 2.02 s; range = 24.2 m; impact velocity = <12.0, −19.8> m/s, with speed 23.2 m/s.
Why it works: Vertically, 20.0 = ½gt2 gives t = 2.02 s. Horizontally, x = (12.0)(2.02) = 24.2 m. At impact, vy = −gt = −19.8 m/s while vx remains 12.0 m/s.
Ready to check your understanding?
You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?
Confidence Check
You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.
QUICK CHECK 1
Horizontal–Vertical Independence
Choose the statement that matches the ideal model.
For r(t) = <2t3, 5t2, 7> m, what are v(1.00 s) and a(1.00 s)?
Reveal Answers
v(1.00 s) = <6, 10, 0> m/s; a(1.00 s) = <12, 10, 0> m/s2.
Why it works: v(t) = <6t2, 10t, 0>; a(t) = <12t, 10, 0>. Evaluate each component at 1.00 s.
QUICK CHECK 2
Apex Check
Identify what is zero and what is not.
At the top of an ideal angled projectile path, which of these are zero: total velocity, vertical velocity, horizontal velocity, acceleration?
Reveal Answers
Only the vertical velocity is zero. Horizontal velocity and downward acceleration are nonzero, so total velocity is nonzero.
Why it works: The apex is a turning point for vertical motion only.
QUICK CHECK 3
Read the Arrows
Use the physical directions, not memorized wording.
A projectile is moving down and to the right. Which way do its velocity and acceleration vectors point?
Reveal Answers
Velocity points down and to the right, tangent to the path; acceleration points straight downward.
Why it works: Velocity describes the current direction of motion. Gravity determines the acceleration direction.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Separate the Motions
Treat horizontal and vertical motion independently, but remember that both components describe the same object at the same time.
KEY TAKEAWAY 2
The Apex Does Not Turn Gravity Off
At the highest point only vertical velocity is momentarily zero; horizontal velocity remains and acceleration is still downward.
KEY TAKEAWAY 3
Use the Model Deliberately
Ideal projectile motion is an approximation that neglects air effects and assumes nearly constant gravitational acceleration.
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