FULL REVIEW

Full Review: Projectile Motion — Calculus-Based

Review the essential ideas, relationships, and problem-solving tools for projectile motion.

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TIME

45–60 minutes

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BEST FOR

A complete topic review

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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.

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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: Calculus-based introductory university physics

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✓ Before starting homework

✓ Before a quiz or exam

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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.

①

Warm-Up Check

Activate prior knowledge.

②

Core Concepts

Review the essential ideas.

③

Guided Practice

Apply what you learned.

④

Confidence Check

Confirm your understanding.

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

Write the vector position function and differentiate component by component.

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 a derivative condition to locate the apex.

A projectile has velocity components v_x = 12.0 m/s and v_y = 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

v_x = 12.0 m/s; v_y = 3.30 m/s; speed = 12.4 m/s; direction = 15.4° above +x.

Why it works: Horizontal acceleration is zero, so v_x stays 12.0 m/s. Vertically, v_y = 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

Eliminate the parameter t to obtain the path equation.

Given x = v₀ cosθ · t and y = y₀ + v₀ sinθ · t − ½gt², what type of function y(x) results?

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 a_x = 0 and a_y = −g. Horizontal velocity stays constant while the vertical velocity changes uniformly.

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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?

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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

Projectile Motion from Vector Differential Equations

The ideal projectile model follows from the vector equation d²r/dt² = <0, −g>. Integrating with initial conditions produces the component velocity and position functions. This formulation makes the independence of x and y explicit and shows that the common parameter t connects the components.

r(t) = <x(t), y(t), z(t)>; Δr = r₂ − r₁; v_avg = Δr/Δt; v(t) = dr/dt; a(t) = dv/dt; speed = √(v_x² + v_y² + v_z²).

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 Equation, Extrema, and Range

Eliminating time from the parametric equations gives the parabolic trajectory y(x). The apex follows from dy/dt = 0 or dy/dx = 0. Landing times are roots of the vertical coordinate equation. For equal-height launches, the nonzero root yields the familiar time and range formulas; unequal heights require the general root.

v_0,x = v₀ cos θ; v_0,y = v₀ sin θ; x = x₀ + v_0,xt; y = y₀ + v_0,yt − ½gt²; v_y = v_0,y − gt. For equal heights: T = 2v_0,y/g, R = v₀² sin(2θ)/g, H = v_0,y²/(2g).

Example: A ball launched at 20.0 m/s and 35.0° from a height of 1.80 m has v_0,x = 16.4 m/s and v_0,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 parabolic y(x) trajectory is a consequence of constant downward acceleration and zero horizontal acceleration; it is not a universal property of motion through a fluid.

KEY CONCEPT 3

Velocity Geometry, Impact State, and Inverse Problems

The tangent to r(t) is the velocity vector. At any event time, speed is |v| and direction follows from the component ratio. In inverse problems, specified target coordinates impose simultaneous constraints on x(t) and y(t). Solving them can reveal required launch parameters or whether a target is reachable under the ideal model.

a_rad = v²/r = ω²r (inward); a_tan = dv/dt; |a| = √(a_rad² + a_tan²); v_A/C = v_A/B + v_B/C.

Example: A ball launched at 18.0 m/s and 40.0° from 5.00 m above the ground has v_0x = 13.8 m/s and v_0y = 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: Use atan2(vᵧ, vₓ) or explicit quadrant reasoning for velocity direction; a simple arctangent can lose quadrant information.

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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?

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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

Integrate the acceleration vector and apply initial conditions.

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

Find the positive landing root, then evaluate the velocity vector there.

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 v_0,x = 16.4 m/s and v_0,y = 11.5 m/s. The positive root of 0 = 1.80 + 11.5t − 4.90t² is 2.49 s. Then x = v_0,xt. The rise above launch is v_0,y²/(2g). At impact, v_y = v_0,y − gt = −12.9 m/s.

PRACTICE 3

Independent Problem

Eliminate time and identify the curvature of the trajectory.

A projectile is launched from the origin with speed v₀ at angle θ. Derive y(x) for the ideal model.

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, v_y = −gt = −19.8 m/s while v_x 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?

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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

Differentiate the Trajectory

Relate the slope of the path to the velocity components.

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

Classify the Apex

Use first and second derivatives.

Why does vᵧ = 0 identify a maximum rather than a minimum for an ideal upward-launched projectile?

Reveal Answers

Because d²y/dt² = −g < 0 at all times.

Why it works: The first derivative vanishes at the apex and the negative second derivative makes y(t) concave downward, so the stationary point is a maximum.

QUICK CHECK 3

Parameter Elimination Check

Identify the source of the quadratic term.

Why does eliminating t from ideal projectile x(t) and y(t) produce a quadratic in x?

Reveal Answers

Because x is linear in t while y contains a t² term; substituting t ∝ x converts t² into x².

Why it works: Zero horizontal acceleration makes x(t) linear. Constant vertical acceleration makes y(t) quadratic. Their combination yields a parabola.