FULL REVIEW
Full Review: Projectile Motion — Algebra-Based
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: Algebra-based introductory college 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
Resolve the launch velocity into horizontal and vertical components.
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
Use the shared-time idea to update both velocity components.
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
Recognize when the equal-height shortcuts are valid.
A projectile launches and lands at the same height. Which quantities can be found directly from the standard equal-height formulas?
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
Component Equations and Shared Time
Resolve the initial velocity before applying kinematics. In the ideal model, aₓ = 0 and aᵧ = −g, so x motion uses constant velocity while y motion uses constant acceleration. Both axes must use the same elapsed time. Often the vertical equation determines time and the horizontal equation then gives range.
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
Range, Height, and Unequal-Height Launches
When launch and landing heights are equal, symmetry gives useful shortcuts. When they differ, return to the general vertical position equation and solve the resulting quadratic for the physically meaningful time. Maximum height follows from vᵧ = 0 or from the vertical kinematic relation that eliminates time.
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: The range formula R = v₀² sin(2θ)/g is not valid when launch and landing heights differ.
KEY CONCEPT 3
Impact Velocity, Horizontal Launches, and Targeting
For a horizontal launch, v₀y = 0 and the drop determines flight time. For any projectile, compute impact velocity from its components at the impact time. Targeting problems work backward: impose the target coordinates and solve the component equations for the unknown time, speed, or angle.
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: Keep the sign of vᵧ until the final interpretation. A negative vertical component means downward when +y is upward.
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 the equal-height formulas and verify the result with components.
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
Solve the vertical quadratic first, then use the same time horizontally.
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
Use the vertical drop to get time, then compute the impact vector.
A package leaves a horizontal conveyor at 18.0 m/s from a height of 45.0 m. Find the landing time, horizontal distance, and impact speed.
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
Choose the Correct Model
Identify whether a shortcut or the general equations are required.
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 Velocity
Evaluate the components at the top of the path.
A projectile is launched at 25.0 m/s at 50.0°. What are vₓ, vᵧ, and acceleration at the apex?
Reveal Answers
vₓ = 16.1 m/s; vᵧ = 0; acceleration = 9.80 m/s² downward.
Why it works: Horizontal velocity remains v₀ cos50° = 16.1 m/s. The vertical velocity reaches zero at the turning point, but gravity remains unchanged.
QUICK CHECK 3
Impact Vector
Do not report speed alone.
Why is an impact velocity not fully described by a single positive number?
Reveal Answers
Because velocity is a vector; its horizontal and vertical components or its speed and direction are required.
Why it works: The sign and direction of vᵧ distinguish rising from falling motion even when the speed magnitude is the same.
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
Resolve First
Convert the launch speed and angle into v₀x and v₀y before using one-dimensional kinematics.
KEY TAKEAWAY 2
One Time Connects Both Axes
Horizontal and vertical equations are solved separately but must describe the same elapsed time.
KEY TAKEAWAY 3
Shortcuts Have Conditions
Equal-height formulas are efficient only when their assumptions match the physical situation.
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