5aProjectile Motion
Analyze the parabolic flight of projectiles by separating horizontal and vertical motion components.
Projectile motion is the foundation for understanding ballistics, sports physics, and any object moving through the air — from a basketball shot to a rocket trajectory.
Lesson Overview
Projectile motion describes the curved path of an object launched into the air under the influence of gravity alone. The key insight is that horizontal and vertical motions are completely independent: horizontally the object moves at constant velocity (aₓ = 0), while vertically it accelerates downward at g = 9.8 m/s². Together these produce a parabolic trajectory.
Key Concepts
Independence of Motion
Horizontal and vertical components are analyzed separately and do not affect each other.
Horizontal Component
vₓ = v₀cosθ; constant throughout flight (aₓ = 0).
Vertical Component
v_y = v₀sinθ − gt; accelerates downward at g = 9.8 m/s².
Parabolic Trajectory
The combination of constant horizontal speed and vertical free-fall produces a parabola.
Range Formula
R = v₀²sin2θ / g — maximum range occurs at θ = 45°.
Time of Flight
T = 2v₀sinθ / g — total time the projectile is in the air.
A ball is launched horizontally at 20 m/s from a cliff 45 m high. How long does it take to hit the ground?
Using the previous problem, how far from the base of the cliff does the ball land?
A soccer ball is kicked at 25 m/s at 30° above the horizontal. Find the maximum height.
Find the range of the soccer ball from Worked Example 3.
At what launch angle is the horizontal range of a projectile maximized? Justify using the range formula.
External Supplemental Resource
Guided Practice Video: Projectile Motion
Review projectile motion formulas, horizontal and vertical components, and worked examples before completing the guided practice problems below.
Video by The Organic Chemistry Tutor on YouTube
Watch on YouTube ↗A stone is thrown horizontally at 15 m/s from a bridge 20 m above a river. How long until it hits the water?
Hint: Only gravity acts vertically; use h = ½gt² with h = 20 m.
A projectile is launched at 40 m/s at 60°. What are the initial horizontal and vertical velocity components?
Hint: Use vₓ = v₀cosθ and v_y = v₀sinθ.
A ball is launched at 30 m/s at 45°. What is its speed at the highest point of its trajectory?
Hint: At the peak, the vertical velocity is zero. What remains?
Two projectiles are launched at the same speed but at 30° and 60°. Compare their ranges.
Hint: Use R = v₀²sin2θ/g. What is sin(60°) and sin(120°)?
A cannonball is fired at 50 m/s at 37°. Find the time of flight.
Hint: Use T = 2v₀sinθ/g.
Key Vocabulary
Projectile
Any object launched into the air that moves under gravity alone (no thrust).
Example: A kicked football, a thrown baseball, or a bullet in flight.
Trajectory
The curved path followed by a projectile through space.
Example: A ball thrown at an angle follows a parabolic trajectory.
Range
The horizontal distance traveled by a projectile from launch to landing.
Example: R = v₀²sin2θ/g gives the range on level ground.
Time of Flight
The total time a projectile remains in the air.
Example: T = 2v₀sinθ/g for a projectile launched and landing at the same height.
Interactive Practice — 5 Questions
Which component of projectile motion has zero acceleration (ignoring air resistance)?
A projectile is launched at 20 m/s at 45°. What is its horizontal velocity component?
At the highest point of its trajectory, a projectile's vertical velocity is:
The range of a projectile is maximized at a launch angle of:
A ball is thrown horizontally from a height. Which statement is true?
Independent Practice
A ball rolls off a table at 3 m/s and the table is 1.25 m high. Find (a) the time to hit the floor and (b) the horizontal distance from the table edge.
A projectile is launched at 50 m/s at 37°. Find (a) the maximum height, (b) the time of flight, and (c) the range.
Explain why two projectiles launched at complementary angles (e.g., 30° and 60°) have the same range.
A ball is thrown at 15 m/s horizontally from a cliff. It lands 45 m from the base. How high is the cliff?
★ A stunt car drives off a ramp angled at 20° at 30 m/s. The landing zone is 5 m below the ramp exit. Find the horizontal distance to the landing point. (Hint: set up the vertical equation y = v_y0·t − ½gt² = −5 and solve for t.)
ChallengeCommon Mistakes
Using the full launch speed v₀ in vertical equations instead of the vertical component v_y0 = v₀sinθ.
Always resolve v₀ into components first: vₓ = v₀cosθ and v_y0 = v₀sinθ, then use each component in its own equation.
Assuming the projectile stops moving horizontally at the peak.
Horizontal velocity is constant throughout the entire flight; only vertical velocity is zero at the peak.
Forgetting that g is negative (downward) when setting up vertical equations.
Take upward as positive; then a_y = −9.8 m/s² and use y = v_y0·t − ½gt².
Math Tips
Always start by resolving the launch velocity into components: vₓ = v₀cosθ, v_y0 = v₀sinθ. Then treat x and y motions as two separate 1-D kinematics problems linked only by time t.
The range formula R = v₀²sin2θ/g only works when launch and landing heights are equal. For unequal heights, solve the vertical equation for t first, then find x = vₓ·t.