Projectile Motion Examples and Problem Solving
What Projectile Motion Actually Is
Projectile motion is the curved path an object follows when you throw it, launch it, or drop it near Earth's surface. The only force acting on it (ignoring air resistance) is gravity pulling it downward.
This means the motion splits into two independent parts: horizontal and vertical. They don't affect each other. That's the whole trick to solving these problems.
The Core Equations You Need
Before touching any example, memorize these. They're not optional.
- Horizontal: x = v₀ · cos(θ) · t
- Vertical: y = v₀ · sin(θ) · t - ½gt²
- Velocity magnitude: v = √(vₓ² + vᵧ²)
- Time of flight: T = 2v₀sin(θ)/g
- Maximum height: H = v₀²sin²(θ)/2g
- Range: R = v₀²sin(2θ)/g
g = 9.8 m/s² (or 32 ft/s² if you're using imperial). Pick one and stay consistent.
Example 1: The Basic Cannonball
A cannon fires a ball at 50 m/s at 30° above the horizontal. Find the horizontal distance it travels.
Step 1: Break the initial velocity into components.
- v₀ₓ = 50 · cos(30°) = 50 · 0.866 = 43.3 m/s
- v₀ᵧ = 50 · sin(30°) = 50 · 0.5 = 25 m/s
Step 2: Find total flight time.
The ball goes up, comes back down. Time = 2 × v₀ᵧ / g = 2 × 25 / 9.8 = 5.1 seconds
Step 3: Multiply horizontal velocity by time.
Range = 43.3 × 5.1 = 221 meters
That's it. Three steps. Horizontal velocity stays constant. Vertical motion handles gravity. Multiply them through time.
Example 2: Dropping Something From a Moving Vehicle
A car drives at 20 m/s horizontally when it drops a package. The package hits the ground after 3 seconds. Where does it land?
Horizontal motion doesn't change. The package keeps moving forward at 20 m/s.
Horizontal distance = 20 × 3 = 60 meters
The vertical motion tells you nothing about horizontal distance. It only tells you when the package hits the ground. The car's speed is the package's horizontal speed because nothing slows it horizontally.
Example 3: Maximum Range Problem
You want to throw a ball as far as possible. What angle maximizes distance?
Look at the range equation: R = v₀²sin(2θ)/g
v₀ and g are fixed. Range is maximum when sin(2θ) = 1, which happens when 2θ = 90°, so θ = 45°.
This only works on flat ground with no air resistance. In the real world, baseball players throw at lower angles because spin and air resistance change everything.
Common Mistakes Students Make
- Using the wrong sign for gravity (it's always negative in the vertical equation)
- Confusing velocity components with total velocity
- Trying to use horizontal equations for vertical problems
- Forgetting that time is the same for both horizontal and vertical motion
- Using launch angle in the wrong equation
Quick Reference: Angle Effects
| Angle | Use Case | Characteristics |
|---|---|---|
| 45° | Maximum range on flat ground | Equal horizontal and vertical components |
| 90° | Maximum height only | Object goes straight up and down |
| 30° | Short range, high arc | Longer air time, less horizontal speed |
| 0° | Horizontal motion only | No vertical component at launch |
How to Solve Any Projectile Motion Problem
Step 1: Identify what you know. Write down initial velocity, angle, height, or time. Circle what the problem asks for.
Step 2: Split the initial velocity into horizontal and vertical components using cosine and sine.
Step 3: Solve the vertical part first. This usually gives you time of flight.
Step 4: Plug that time into the horizontal equation.
Step 5: Check that your units make sense. Distance should be in meters. Time should be in seconds.
When to Ignore Air Resistance
In physics class, always. In the real world, never.
For a cannonball, air resistance barely matters. For a feather, it dominates everything. For a basketball, it matters enough to change the optimal angle from 45° to around 42-43°.
The equations you learned assume no air resistance. That's why physics problems use smooth spheres and ignore wind.