Projectile Motion

Motion under gravity in two dimensions

Projectile Motion

Introduction

Projectile motion is 2D motion under the influence of gravity only (ignoring air resistance).

Examples:

Ball thrown at an angle
Cannonball fired from a cannon
Water from a fountain
Long jumper

Key Characteristics

Horizontal motion: constant velocity (no horizontal acceleration) $a_x = 0$ $v_x = v_{0x} = \text{constant}$
Vertical motion: constant acceleration due to gravity $a_y = -g = -9.8 \text{ m/s}^2$
Parabolic path: The trajectory is a parabola
Independence: Horizontal and vertical motions are independent

Initial Velocity Components

If launched at angle $\theta_0$ with initial speed $v_0$ :

$v_{0x} = v_0 \cos \theta_0$ $v_{0y} = v_0 \sin \theta_0$

Equations of Motion

Horizontal Direction (constant velocity)

$x = x_0 + v_{0x}t$ $v_x = v_{0x} = \text{constant}$

Vertical Direction (constant acceleration)

$y = y_0 + v_{0y}t - \frac{1}{2}gt^2$ $v_y = v_{0y} - gt$ $v_y^2 = v_{0y}^2 - 2g(y - y_0)$

Note: Using $-g$ for downward acceleration.

Key Features of Parabolic Trajectory

At the Peak

Vertical velocity: $v_y = 0$ (changes from positive to negative)
Horizontal velocity: $v_x = v_{0x}$ (unchanged)
Total velocity: $v = v_x$ (purely horizontal)
Acceleration: $a = -g$ (always downward, even at peak!)

Symmetry Properties

For projectile launched and landing at same height:

Time up = Time down $t_{up} = t_{down} = \frac{v_{0y}}{g}$
Total time of flight: $t_{total} = \frac{2v_{0y}}{g} = \frac{2v_0 \sin \theta_0}{g}$
Maximum height: $h_{max} = \frac{v_{0y}^2}{2g} = \frac{(v_0 \sin \theta_0)^2}{2g}$
Range (horizontal distance): $R = \frac{v_0^2 \sin(2\theta_0)}{g}$
Landing speed = Launch speed (magnitude)
Landing angle = Launch angle (below horizontal)

Special Cases

Horizontal Launch ( $\theta_0 = 0°$ )

$v_{0x} = v_0$
$v_{0y} = 0$
Time in air: $t = \sqrt{\frac{2h}{g}}$ where $h$ is initial height

Vertical Launch ( $\theta_0 = 90°$ )

$v_{0x} = 0$ (no horizontal motion)
$v_{0y} = v_0$
1D motion only

45° Launch (Maximum Range)

For a given speed $v_0$ , $\theta_0 = 45°$ gives maximum range
$R_{max} = \frac{v_0^2}{g}$

Complementary Angles

$\theta_0$ and $(90° - \theta_0)$ give the same range
Example: 30° and 60° both give same range (but different trajectories!)

Problem-Solving Strategy

Draw a diagram showing trajectory
Set up coordinate system (origin at launch point)
List knowns: $v_0$ , $\theta_0$ , $x_0$ , $y_0$ , etc.
Break $v_0$ into components: $v_{0x}$ and $v_{0y}$
Write separate equations for $x$ and $y$
Use time as the link between horizontal and vertical
Solve step by step

Common Questions

Q: What is the acceleration at the peak? A: $a = -g$ downward. Acceleration is always $-g$ throughout flight!

Q: What is the velocity at the peak? A: $v = v_x = v_{0x}$ (horizontal only, since $v_y = 0$ )

Q: How do you find time to reach maximum height? A: Use $v_y = v_{0y} - gt$ and set $v_y = 0$ : $t = \frac{v_{0y}}{g}$

Q: How do you find range? A: Find total time of flight, then $R = v_x \cdot t_{total}$

Real-World Applications

Sports: basketball, soccer, golf, baseball
Military: ballistics, artillery
Engineering: water fountains, sprinkler systems
Space: satellite trajectories (approximately)

📚 Practice Problems

1Problem 1medium

❓ Question:

A ball is kicked at 20 m/s at an angle of 37° above the horizontal. Use g = 10 m/s². (a) What are the horizontal and vertical components of the initial velocity? (b) What is the maximum height? (c) What is the time of flight? (d) What is the range?

💡 Show Solution

Solution:

Given: v₀ = 20 m/s, θ = 37°, g = 10 m/s² Note: sin 37° ≈ 0.60, cos 37° ≈ 0.80

(a) Initial velocity components: v₀ₓ = v₀ cos θ = 20(0.80) = 16 m/s v₀ᵧ = v₀ sin θ = 20(0.60) = 12 m/s

(b) Maximum height: At max height, vᵧ = 0 Using vᵧ² = v₀ᵧ² - 2gΔy 0 = (12)² - 2(10)Δy 20Δy = 144 Δy = 7.2 m

(c) Time of flight: Using Δy = v₀ᵧt - ½gt² with Δy = 0 (lands at same height) 0 = 12t - 5t² t(12 - 5t) = 0 t = 0 or t = 12/5 = 2.4 s

(d) Range: R = v₀ₓ × t_total = 16 × 2.4 = 38.4 m or 38 m

Alternatively: R = (v₀² sin 2θ)/g = (400 × sin 74°)/10 ≈ 38.4 m

2Problem 2easy

❓ Question:

A ball is kicked at $20$ m/s at an angle of $30°$ above the horizontal. Find the horizontal and vertical components of the initial velocity.

💡 Show Solution

Given:

Initial speed: $v_0 = 20$ m/s
Launch angle: $\theta_0 = 30°$

Find: $v_{0x}$ and $v_{0y}$

Horizontal component: $v_{0x} = v_0 \cos \theta_0$ $v_{0x} = 20 \cos(30°)$ $v_{0x} = 20 \times 0.866$ $v_{0x} \approx 17.3 \text{ m/s}$

Vertical component: $v_{0y} = v_0 \sin \theta_0$ $v_{0y} = 20 \sin(30°)$ $v_{0y} = 20 \times 0.5$ $v_{0y} = 10 \text{ m/s}$

Answers:

Horizontal component: 17.3 m/s
Vertical component: 10 m/s

Check: $\sqrt{17.3^2 + 10^2} = \sqrt{299.29 + 100} \approx 20$ m/s ✓

3Problem 3medium

❓ Question:

💡 Show Solution

Solution:

Given: v₀ = 20 m/s, θ = 37°, g = 10 m/s² Note: sin 37° ≈ 0.60, cos 37° ≈ 0.80

(a) Initial velocity components: v₀ₓ = v₀ cos θ = 20(0.80) = 16 m/s v₀ᵧ = v₀ sin θ = 20(0.60) = 12 m/s

(b) Maximum height: At max height, vᵧ = 0 Using vᵧ² = v₀ᵧ² - 2gΔy 0 = (12)² - 2(10)Δy 20Δy = 144 Δy = 7.2 m

(c) Time of flight: Using Δy = v₀ᵧt - ½gt² with Δy = 0 (lands at same height) 0 = 12t - 5t² t(12 - 5t) = 0 t = 0 or t = 12/5 = 2.4 s

(d) Range: R = v₀ₓ × t_total = 16 × 2.4 = 38.4 m or 38 m

Alternatively: R = (v₀² sin 2θ)/g = (400 × sin 74°)/10 ≈ 38.4 m

4Problem 4hard

❓ Question:

A ball rolls off a table 1.2 m high with a horizontal velocity of 3.0 m/s. (a) How long does it take to hit the ground? (b) How far from the base of the table does it land? (c) What is its velocity (magnitude and direction) just before impact?

💡 Show Solution

Solution:

Given: h = 1.2 m, v₀ₓ = 3.0 m/s, v₀ᵧ = 0 (horizontal launch), g = 10 m/s²

(a) Time to hit ground: Using Δy = v₀ᵧt + ½gt² (taking down as positive) 1.2 = 0 + ½(10)t² 1.2 = 5t² t² = 0.24 t = 0.49 s or 0.5 s

(b) Horizontal distance: x = v₀ₓt = 3.0 × 0.49 = 1.47 m or 1.5 m

(c) Velocity at impact: Horizontal: vₓ = v₀ₓ = 3.0 m/s (constant) Vertical: vᵧ = v₀ᵧ + gt = 0 + 10(0.49) = 4.9 m/s (downward)

Magnitude: v = √(vₓ² + vᵧ²) = √(3.0² + 4.9²) = √(9 + 24) = 5.7 m/s

Direction: θ = tan⁻¹(vᵧ/vₓ) = tan⁻¹(4.9/3.0) = 59° below horizontal

5Problem 5hard

❓ Question:

💡 Show Solution

Solution:

Given: h = 1.2 m, v₀ₓ = 3.0 m/s, v₀ᵧ = 0 (horizontal launch), g = 10 m/s²

(a) Time to hit ground: Using Δy = v₀ᵧt + ½gt² (taking down as positive) 1.2 = 0 + ½(10)t² 1.2 = 5t² t² = 0.24 t = 0.49 s or 0.5 s

(b) Horizontal distance: x = v₀ₓt = 3.0 × 0.49 = 1.47 m or 1.5 m

(c) Velocity at impact: Horizontal: vₓ = v₀ₓ = 3.0 m/s (constant) Vertical: vᵧ = v₀ᵧ + gt = 0 + 10(0.49) = 4.9 m/s (downward)

Magnitude: v = √(vₓ² + vᵧ²) = √(3.0² + 4.9²) = √(9 + 24) = 5.7 m/s

Direction: θ = tan⁻¹(vᵧ/vₓ) = tan⁻¹(4.9/3.0) = 59° below horizontal

6Problem 6medium

❓ Question:

A projectile is launched horizontally from a cliff $80$ m high with initial speed $30$ m/s. How long is it in the air, and how far from the base of the cliff does it land? (Use $g = 10$ m/s²)

💡 Show Solution

Given:

Initial height: $y_0 = 80$ m
Horizontal launch: $\theta_0 = 0°$
Initial speed: $v_0 = 30$ m/s
$v_{0x} = 30$ m/s, $v_{0y} = 0$ m/s
$g = 10$ m/s²

Find:

Time in air
Horizontal distance (range)

Part 1: Time in air

Use vertical motion equation: $y = y_0 + v_{0y}t - \frac{1}{2}gt^2$

At landing, $y = 0$ : $0 = 80 + 0 \cdot t - \frac{1}{2}(10)t^2$ $0 = 80 - 5t^2$ $5t^2 = 80$ $t^2 = 16$ $t = 4 \text{ s}$

Part 2: Horizontal distance

Use horizontal motion (constant velocity): $x = v_{0x} \cdot t$ $x = 30 \times 4$ $x = 120 \text{ m}$

Answers:

Time in air: 4 seconds
Horizontal distance: 120 m from base of cliff

Note: The horizontal speed doesn't affect time in air—only the height does!

7Problem 7hard

❓ Question:

A soccer ball is kicked at $25$ m/s at an angle of $37°$ above horizontal. Find: (a) the maximum height reached, (b) the time of flight, and (c) the range. Use $g = 10$ m/s² and $\sin(37°) \approx 0.6$ , $\cos(37°) \approx 0.8$ .

💡 Show Solution

Given:

Initial speed: $v_0 = 25$ m/s
Launch angle: $\theta_0 = 37°$
$g = 10$ m/s²

Step 1: Find initial velocity components $v_{0x} = v_0 \cos(37°) = 25 \times 0.8 = 20 \text{ m/s}$ $v_{0y} = v_0 \sin(37°) = 25 \times 0.6 = 15 \text{ m/s}$

Part (a): Maximum height

Use $v_y^2 = v_{0y}^2 - 2gh$ where $v_y = 0$ at max height: $0 = (15)^2 - 2(10)h$ $0 = 225 - 20h$ $20h = 225$ $h = 11.25 \text{ m}$

Alternative: $h_{max} = \frac{v_{0y}^2}{2g} = \frac{225}{20} = 11.25$ m

Part (b): Time of flight

Time to peak: $t_{up} = \frac{v_{0y}}{g} = \frac{15}{10} = 1.5$ s

Total time (up and down): $t_{total} = 2 \times 1.5 = 3.0$ s

Part (c): Range

Horizontal distance traveled: $R = v_{0x} \times t_{total}$ $R = 20 \times 3.0$ $R = 60 \text{ m}$

Alternative formula: $R = \frac{v_0^2 \sin(2\theta_0)}{g} = \frac{625 \times 2 \times 0.6 \times 0.8}{10} = \frac{600}{10} = 60$ m

Answers:

(a) Maximum height: 11.25 m
(b) Time of flight: 3.0 s
(c) Range: 60 m

Check: At landing, $v_y = -15$ m/s (same magnitude as launch, opposite direction) ✓

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Projectile Motion

Projectile Motion

Introduction

Key Characteristics

Initial Velocity Components

Equations of Motion

Horizontal Direction (constant velocity)

Vertical Direction (constant acceleration)

Key Features of Parabolic Trajectory

At the Peak

Symmetry Properties

Special Cases

Horizontal Launch (θ0=0°\theta_0 = 0°θ0​=0°)

Vertical Launch (θ0=90°\theta_0 = 90°θ0​=90°)

45° Launch (Maximum Range)

Complementary Angles

Problem-Solving Strategy

Common Questions

Real-World Applications

📚 Practice Problems

1Problem 1medium

❓ Question:

2Problem 2easy

❓ Question:

3Problem 3medium

❓ Question:

4Problem 4hard

❓ Question:

5Problem 5hard

❓ Question:

6Problem 6medium

❓ Question:

7Problem 7hard

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Horizontal Launch ( $\theta_0 = 0°$ )

Vertical Launch ( $\theta_0 = 90°$ )