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:

1. Time up = Time down $t_{up} = t_{down} = \frac{v_{0y}}{g}$

2. Total time of flight: $t_{total} = \frac{2v_{0y}}{g} = \frac{2v_0 \sin \theta_0}{g}$

3. Maximum height: $h_{max} = \frac{v_{0y}^2}{2g} = \frac{(v_0 \sin \theta_0)^2}{2g}$

4. Range (horizontal distance): $R = \frac{v_0^2 \sin(2\theta_0)}{g}$

5. Landing speed = Launch speed (magnitude)

6. 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 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

1. Draw a diagram showing trajectory
2. Set up coordinate system (origin at launch point)
3. List knowns: $v_0$, $\theta_0$, $x_0$, $y_0$, etc.
4. Break $v_0$ into components: $v_{0x}$ and $v_{0y}$
5. Write separate equations for $x$ and $y$
6. Use time as the link between horizontal and vertical
7. 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)

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 ✓

Find:

1. Time in air
2. 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!

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) ✓