🚀Projectile Motion
The Key Insight
Projectile motion is any motion where an object moves through the air under the influence of gravity alone (no engine, no air resistance). The key insight that unlocks all projectile problems is:
Horizontal and vertical motion are completely independent.
- Horizontal: constant velocity (no force acting horizontally) - Vertical: constant acceleration downward at g = 9.8 m/s² ≈ 10 m/s²
This means you can split any projectile problem into two separate 1D kinematic problems and solve each one independently.
Decomposing Initial Velocity
If a projectile is launched at angle θ with speed v₀:
v₀ₓ = v₀ cos θ (horizontal component) v₀ᵧ = v₀ sin θ (vertical component)
Once launched, horizontal velocity stays constant throughout the flight. Vertical velocity changes due to gravity.
At the peak of the trajectory, vertical velocity = 0 (but horizontal velocity is still v₀ₓ!).
Projectile Motion Simulator
Launch the projectile at different angles and initial speeds. Notice how horizontal and vertical motions are independent.
Think About It
If you fire a bullet horizontally from a gun and simultaneously drop a bullet from the same height, which hits the ground first?
✏️ Worked Example
Problem: A ball is thrown horizontally at 15 m/s from a 20 m cliff. How far from the base of the cliff does it land?
📐 Key Equations for Projectile Motion
Treat horizontal and vertical separately:
x = v_0x · t (horizontal, constant velocity)y = v_0yt - (1)/(2)gt² (vertical)v_y = v_0y - gtv_0x = v_0\cosθ, v_0y = v_0\sinθ⚠️ Common Mistakes
Misconception: The horizontal velocity decreases during flight because gravity acts on the projectile.
✓ Correct thinking: Gravity only affects the vertical component. Horizontal velocity stays constant throughout the flight (ignoring air resistance).
Why: This misunderstanding collapses the independence of horizontal and vertical motion, which is the entire foundation of projectile analysis.
Misconception: At the peak of its trajectory, a projectile has zero velocity.
✓ Correct thinking: At the peak, only the vertical velocity is zero. Horizontal velocity is still v₀ cosθ. The total speed at the peak equals v₀ cosθ.
Why: Only the vertical component reverses direction at the peak; the horizontal component is unchanged.
Misconception: A 45° launch angle always gives the maximum range.
✓ Correct thinking: 45° gives maximum range only when the launch and landing heights are equal. If launching from a height (e.g., off a cliff), the optimal angle is less than 45°.
Why: The 45° result is derived assuming symmetric parabolic flight. Asymmetric situations shift the optimum.
📝 Practice Problems
Try these problems. Check your answer when ready.
A ball rolls off a table 1.25 m high at 3 m/s horizontally. How far from the base of the table does it land? (g = 10 m/s²)
A projectile is launched at 30° above horizontal with an initial speed of 20 m/s. Find the horizontal and vertical components of the initial velocity.
v_0x = v_0\cos 30°, v_0y = v_0\sin 30°A soccer ball is kicked with v₀ₓ = 12 m/s and v₀ᵧ = 16 m/s. What is the ball's speed at the peak of its trajectory?
A ball is launched horizontally from a 80 m high cliff at 20 m/s. Find (a) time of flight, (b) horizontal range, (c) speed just before impact. (g = 10 m/s²)
A projectile is launched at 40 m/s at 37° above horizontal (sin37° = 0.6, cos37° = 0.8). Find (a) maximum height, (b) total time of flight, (c) horizontal range. (g = 10 m/s²)
A cannon on top of a 100 m cliff fires a cannonball horizontally. The cannonball must clear a 20 m tall wall located 150 m from the base of the cliff. What minimum horizontal launch speed is required? (g = 10 m/s²)
Two identical balls are launched simultaneously from the same point. Ball A is launched horizontally at 10 m/s; Ball B is dropped straight down. Which hits the ground first, and why?
Finished reading through this lesson?