🚀 Horizontal Launch

Part 1 of 7 — Projectile Motion

A projectile is any object that moves through the air under the influence of gravity alone (ignoring air resistance). The simplest case is a horizontal launch — when an object is launched with only horizontal velocity and no initial vertical velocity.

Horizontal Launch Setup

When a projectile is launched horizontally:

• Initial horizontal velocity: $v_{0x} = v_0$
• Initial vertical velocity: $v_{0y} = 0$
• Horizontal acceleration: $a_x = 0$
• Vertical acceleration: $a_y = -g$ (downward)

Equations of Motion

Path Shape

The trajectory is a parabola that curves downward. The horizontal spacing remains even (constant $v_x$), while the vertical drops increase (accelerating due to gravity).

Real-World Examples of Horizontal Launch

• A ball rolling off a table
• A package dropped from a moving airplane
• Water flowing over a waterfall
• A bullet fired from a horizontal gun

In all these cases, the object starts with horizontal velocity and zero vertical velocity. Gravity then pulls it downward while it continues forward at constant speed.

Key Insight

The time to fall depends only on the height, not on how fast the object moves horizontally. A ball dropped from a table and a ball launched at 100 m/s from the same table both hit the ground at the same time!

Concept Check 🎯

Horizontal Launch Calculations 🧮

A ball is launched horizontally at 10 m/s from the top of a 20 m cliff. Use $g = 10$ m/s².

1. Time to reach the ground (in seconds)

2. Horizontal distance from the cliff base (in meters)

3. Vertical velocity at impact, magnitude (in m/s)

4. Total speed at impact (in m/s, round to nearest whole number)

Conceptual Review 🔍

Exit Quiz ✅

📐 Angled Launch

Part 2 of 7 — Projectile Motion

Most projectiles aren't launched horizontally — they're launched at an angle. This lesson covers how to decompose the initial velocity into components and solve the full 2D problem.

Decomposing the Launch Velocity

When a projectile is launched at speed $v_0$ at angle $\theta$ above the horizontal:

$v_{0x} = v_0\cos\theta$

$v_{0y} = v_0\sin\theta$

Complete Equations

Special Cases

• $\theta = 0°$: Horizontal launch ($v_{0y} = 0$)
• $\theta = 90°$: Straight up ($v_{0x} = 0$)
• $\theta = 45°$: Equal components ($v_{0x} = v_{0y}$)

Quick Trig Reference

For AP Physics problems, you should know these common values:

Memory trick: 37° and 53° are complementary and use the 3-4-5 ratio scaled by $\frac{1}{5}$.

Component Decomposition Practice 🎯

Angled Launch Problem 🧮

A football is kicked at 25 m/s at 53° above horizontal from ground level. Use $g = 10$ m/s², $\cos 53° = 0.6$, $\sin 53° = 0.8$.

1. $v_{0x}$ (in m/s)

2. $v_{0y}$ (in m/s)

3. Velocity in the $y$-direction after 2 seconds (in m/s)

4. Position in the $y$-direction after 2 seconds (in meters)

Conceptual Understanding 🔍

Exit Problem ✅

A golf ball is hit at 50 m/s at 37° above horizontal. Use $g = 10$ m/s², $\cos 37° = 0.8$, $\sin 37° = 0.6$.

1. $v_{0x}$ (in m/s)

2. $v_{0y}$ (in m/s)

3. Check: $\sqrt{v_{0x}^2 + v_{0y}^2}$ should equal the launch speed (in m/s)

⏱️ Time of Flight and Range

Part 3 of 7 — Projectile Motion

Two of the most important quantities for any projectile are its time of flight (how long it stays in the air) and its range (how far it goes horizontally). This lesson derives and applies these formulas.

Time of Flight

For a projectile launched from and landing at the same height (ground to ground):

At landing, $\Delta y = 0$:

$0 = v_{0y}t - \frac{1}{2}gt^2 = t\left(v_{0y} - \frac{1}{2}gt\right)$

This gives $t = 0$ (launch) or:

$\boxed{t_{flight} = \frac{2v_{0y}}{g} = \frac{2v_0\sin\theta}{g}}$

Key Observations

• Time of flight depends on $v_{0y}$ (the vertical component) only
• Greater launch angle → greater $v_{0y}$ → longer time in air
• Greater launch speed → longer time in air
• $g$ in the denominator: stronger gravity → shorter flight

Range

The range is the horizontal distance traveled during the flight time:

$R = v_{0x} \cdot t_{flight} = v_0\cos\theta \cdot \frac{2v_0\sin\theta}{g}$

Using the identity $2\sin\theta\cos\theta = \sin 2\theta$:

$\boxed{R = \frac{v_0^2\sin 2\theta}{g}}$

Key Observations

• $R$ is maximized when $\sin 2\theta = 1$, which means $2\theta = 90°$, so $\theta = 45°$
• $R \propto v_0^2$: doubling the speed quadruples the range
• Complementary angles ($\theta$ and $90° - \theta$) give the same range
  • Example: 30° and 60° both give $\sin 60° = \sin 120°$

Range at 45°

$R_{max} = \frac{v_0^2}{g}$

Concept Check 🎯

Time and Range Calculations 🧮

A ball is launched from ground level at 30 m/s at 53° above horizontal. Use $g = 10$ m/s², $\cos 53° = 0.6$, $\sin 53° = 0.8$.

1. $v_{0y}$ (in m/s)

2. Time of flight (in seconds)

3. $v_{0x}$ (in m/s)

4. Range (in meters)

Round all answers to 3 significant figures.

Comparing Launch Angles 🔍

Exit Problem ✅

A projectile is launched at 20 m/s at 45° from ground level. Use $g = 10$ m/s², $\sin 45° \approx 0.707$.

1. Time of flight (in seconds, round to 1 decimal)

2. Range using $R = \frac{v_0^2\sin 2\theta}{g}$ (in meters)

Round all answers to 3 significant figures.

⬆️ Maximum Height

Part 4 of 7 — Projectile Motion

The maximum height of a projectile is the highest point in its trajectory. At this point, the vertical velocity is momentarily zero. Let's derive the formula and practice using it.

Deriving the Maximum Height Formula

At the maximum height, $v_y = 0$. Using $v_y^2 = v_{0y}^2 - 2g\Delta y$:

$0 = v_{0y}^2 - 2g h_{max}$

$\boxed{h_{max} = \frac{v_{0y}^2}{2g} = \frac{(v_0\sin\theta)^2}{2g} = \frac{v_0^2\sin^2\theta}{2g}}$

Time to Reach Maximum Height

Using $v_y = v_{0y} - gt_{up}$ with $v_y = 0$:

$t_{up} = \frac{v_{0y}}{g} = \frac{v_0\sin\theta}{g}$

Note: This is exactly half the total flight time for ground-to-ground launches!

$t_{up} = \frac{t_{flight}}{2}$

Key Observations

• $h_{max}$ depends on $v_{0y}$ (vertical component) only
• $h_{max} \propto v_{0y}^2$: doubling $v_{0y}$ quadruples the max height
• For a given speed, $h_{max}$ is greatest at $\theta = 90°$ (straight up)

Concept Check 🎯

Maximum Height Calculations 🧮

A ball is launched at 50 m/s at 37° above horizontal. Use $g = 10$ m/s², $\sin 37° = 0.6$.

1. $v_{0y}$ (in m/s)

2. Maximum height (in meters)

3. Time to reach maximum height (in seconds)

Comparing Heights at Different Angles 🔍

All launches are at the same speed $v_0$.

Challenge Problem 🏆

A ball is launched from ground level and reaches a maximum height of 31.25 m. The total time of flight is 5 s. Use $g = 10$ m/s².

1. $v_{0y}$ (in m/s)

2. Total time of flight from $t = 2v_{0y}/g$ confirms what value? (in seconds)

3. What was the launch speed if $\theta = 53°$? ($\sin 53° = 0.8$) (in m/s)

Round all answers to 3 significant figures.

Exit Quiz ✅

🪞 Symmetry of the Trajectory

Part 5 of 7 — Projectile Motion

Projectile trajectories have beautiful symmetry properties. Understanding these symmetries lets you solve problems faster and check your answers more easily.

Symmetric Properties (Ground-to-Ground)

For a projectile launched from and landing at the same height:

Time Symmetry

• Time up = Time down: $t_{up} = t_{down} = \frac{v_{0y}}{g}$
• Total time: $t_{flight} = 2t_{up}$
• The projectile reaches max height at exactly halfway through the flight

Speed Symmetry

• The speed at any height on the way up = speed at the same height on the way down
• Landing speed = Launch speed: $v_{final} = v_0$
• The vertical velocity at landing has the same magnitude but opposite direction as at launch: $v_{y,final} = -v_{0y}$

Trajectory Symmetry

• The path is a symmetric parabola about the vertical line through the peak
• The horizontal distance to the peak = half the range

Velocity at Key Points

Consider a projectile launched at speed $v_0$ at angle $\theta$:

Important Notes

• $v_x$ never changes — it's the same at every point
• Only $v_y$ changes, and it changes symmetrically
• The minimum speed occurs at the top of the arc
• The projectile hits the ground at the same speed it was launched (for flat ground)

Symmetry Concept Check 🎯

Using Symmetry to Solve Problems 🧮

A ball is launched from the ground at 20 m/s at 30° above horizontal. Use $g = 10$ m/s², $\sin 30° = 0.5$, $\cos 30° \approx 0.866$.

1. Time to reach maximum height (in seconds)

2. Maximum height (in meters)

3. Speed at maximum height (in m/s, round to 1 decimal)

4. $v_y$ at landing (in m/s, include sign)

Round all answers to 3 significant figures.

Symmetry Deep Dive 🔍

⚠️ When Symmetry Breaks

Symmetry applies only when the projectile lands at the same height as it was launched. Symmetry breaks when:

• Launching from a cliff (landing at a different height)
• Landing on an elevated platform
• Air resistance is significant

In these cases, you must solve the full kinematic equations without relying on symmetry shortcuts. The landing speed will not equal the launch speed, and the time up will not equal the time down.

Exit Quiz ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Projectile Motion

This workshop takes you through complete projectile motion problems from start to finish. We'll practice both horizontal and angled launches, including problems from elevated positions.

The Complete Method

Step 1: Identify the Launch Type

• Horizontal launch: $v_{0y} = 0$, $v_{0x} = v_0$
• Angled launch: $v_{0x} = v_0\cos\theta$, $v_{0y} = v_0\sin\theta$

Step 2: Set Up Two DVAT Tables

Step 3: Solve Vertical First (Usually)

• Find time of flight from vertical equation
• Use time in horizontal equation for range

Step 4: Find Final Velocities If Needed

• $v_x = v_{0x}$ (always)
• $v_y = v_{0y} - gt$
• $v = \sqrt{v_x^2 + v_y^2}$

Problem 1 — Horizontal Launch from a Building 🏢

A stone is thrown horizontally at 15 m/s from the roof of a 45 m tall building. Use $g = 10$ m/s².

1. Time to hit the ground (in seconds)

2. Horizontal distance from the building (in meters)

3. Final speed at impact (in m/s, round to nearest whole number)

Problem 2 — Ground-to-Ground Angled Launch ⚽

A soccer ball is kicked from ground level at 40 m/s at 53° above horizontal. Use $g = 10$ m/s², $\cos 53° = 0.6$, $\sin 53° = 0.8$.

1. Maximum height (in meters)

2. Total time of flight (in seconds)

3. Range (in meters)

4. Speed at maximum height (in m/s)

Round all answers to 3 significant figures.

Problem 3 — Conceptual Challenges 🧠

Problem 4 — Quick Analysis 🔍

Problem 5 — Cliff Drop 🪂

A ball is thrown at 30 m/s horizontally from the top of a 125 m cliff. Use $g = 10$ m/s².

1. Time to hit the ground (in seconds)

2. Range (in meters)

Exit Quiz ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Projectile Motion

This is the final lesson of the Projectile Motion unit. We'll review everything — horizontal launches, angled launches, time of flight, range, maximum height, and symmetry — with AP-level questions.

Complete Projectile Motion Summary

Setup

• $v_{0x} = v_0\cos\theta$, $v_{0y} = v_0\sin\theta$
• $a_x = 0$, $a_y = -g$

Key Formulas (Ground-to-Ground)

Symmetry (Ground-to-Ground Only)

• $t_{up} = t_{down}$
• Landing speed = launch speed
• Complementary angles give equal range
• Peak occurs at $t_{flight}/2$ and $R/2$

AP-Style Multiple Choice — Set 1 🎯

AP Calculation Problem 1 🧮

A projectile is launched from ground level at 100 m/s at 53° above horizontal. Use $g = 10$ m/s², $\cos 53° = 0.6$, $\sin 53° = 0.8$.

1. $v_{0x}$ (in m/s)

2. $v_{0y}$ (in m/s)

3. Time of flight (in seconds)

4. Maximum height (in meters)

5. Range (in meters)

Conceptual Mastery 🔍

AP-Style Multiple Choice — Set 2 🎯

AP Free-Response Style 📝

A stunt driver launches a car horizontally off a 20 m high ramp at 25 m/s, aiming to clear a 50 m wide canyon. Use $g = 10$ m/s².

1. Time to fall 20 m (in seconds)

2. Horizontal distance traveled in that time (in meters)

3. Does the car clear the canyon? (enter yes or no)

Final Assessment ✅