Projectile Motion

9.1 What Makes Motion a Projectile

A projectile is any object that is launched into the air and then moves under the influence of gravity alone, with no propulsion. Once the object leaves the launching mechanism — whether a hand, a cannon, or a ramp — it becomes a projectile. The word “projectile” does not imply a weapon; it applies to a thrown baseball, a water droplet from a fountain, a long-jumper in the air, or a package dropped from an airplane.

The defining feature of ideal projectile motion is that the only force acting on the object after launch is gravity. This means the vertical acceleration is a constant 9.81 m/s², downward if upward is chosen as positive, and the horizontal acceleration is exactly zero. These two facts, taken together, completely determine the trajectory.

9.2 Assumptions of Ideal Projectile Motion

The analysis in this chapter rests on three idealizations. State them explicitly at the start of every projectile problem.

• No air resistance: the only force after launch is gravity. This is accurate for dense, compact objects at moderate speeds and breaks down for light, wide, or fast objects.
• Constant gravitational acceleration: g = 9.81 m/s² downward, uniform everywhere during the flight. This is valid as long as the height change is much smaller than Earth’s radius, about 6400 km.
• Flat Earth approximation: the curvature of Earth’s surface is negligible over the range of the projectile. This is valid for ranges much less than 8 km.

9.3 Horizontal and Vertical Motion Are Independent

The foundational principle of projectile analysis is that the horizontal and vertical components of motion are completely independent. Each obeys its own set of kinematic equations. They share only one thing: time. The time elapsed is the same for both components.

This independence follows directly from Newton’s second law applied separately to each direction. Gravity acts only downward; it has zero horizontal component. No force acts horizontally under the assumptions of ideal projectile motion, so there is zero horizontal acceleration. The two directions do not communicate with each other.

9.4 Launch Velocity and Vector Components

The launch velocity is a vector with magnitude v_i, the launch speed, and direction θ above the horizontal. As in Chapter 4, we decompose it into horizontal and vertical components:

v_{x0} = v_i cosθ

v_{y0} = v_i sinθ

These components are the starting values for the two independent kinematic analyses. The horizontal component v_{x0} stays constant throughout the flight. The vertical component v_{y0} changes every instant due to gravitational acceleration.

9.5 Horizontal Motion in Projectile Problems

Because horizontal acceleration is zero, horizontal motion is uniform: constant velocity throughout the flight. The only equation needed for the horizontal direction is:

x = v_{x0}t, where v_{x0} = v_i cosθ

The horizontal velocity v_{x0} never changes. It is the same at launch, at the peak, and just before landing. There is no horizontal force to speed it up or slow it down under the ideal assumptions.

9.6 Vertical Motion in Projectile Problems

Vertical motion is free fall: constant downward acceleration g = 9.81 m/s². Using upward as positive, a_y = −9.81 m/s². The kinematic equations from Chapter 8 apply directly:

v_y = v_{y0} + a_y t

y = v_{y0}t + ½a_y t²

v_y² = v_{y0}² + 2a_y Δy

The vertical and horizontal analyses share time t. Finding t from one component and substituting into the other is the most common technique for linking the two equations.

9.7 Time of Flight

The time of flight is the total time the projectile is in the air from launch to landing. For a projectile launched from and returning to the same height, time of flight is found by setting y = 0 in the vertical kinematic equation and solving for t.

0 = v_{y0}t + ½a_y t² = t(v_{y0} + ½a_y t)

This equation has two solutions: t = 0, the launch instant, and t = 2v_{y0}/g, the landing instant. The time of flight is:

T = 2v_{y0}/g = 2v_i sinθ/g

9.8 Maximum Height

Maximum height is reached when the vertical velocity is zero, v_y = 0. At that instant, the projectile is moving purely horizontally. Using the vertical kinematic equation without time:

0 = v_{y0}² + 2a_yH

H = v_{y0}²/(2g) = (v_i sinθ)²/(2g)

The time to reach maximum height is exactly half the total time of flight for symmetric trajectories:

t_{peak} = v_{y0}/g

9.9 Horizontal Range

The horizontal range R is the total horizontal distance from launch to landing. It is found by substituting the time of flight into the horizontal equation:

R = v_{x0}T = v_i cosθ(2v_i sinθ/g)

R = v_i² sin(2θ)/g

9.10 Launches from Elevated Positions

When a projectile is launched from a height h above the landing point, or below it, the simple formulas for time of flight, maximum height, and range no longer apply directly. Instead, set up the vertical equation with the correct displacement and solve the quadratic.

9.11 Symmetry in Projectile Motion

For a projectile launched and landing at the same height, the trajectory is perfectly symmetric about the peak. This symmetry produces several useful relationships:

• The time to reach the peak equals the time to fall from the peak back to launch height: t_{up} = t_{down} = T/2.
• The speed at any point on the way up equals the speed at the corresponding point on the way down at the same height.
• The launch angle below the horizontal at impact equals the launch angle above the horizontal at launch, in magnitude.
• At the peak, v_y = 0 and v_x = v_{x0}; the speed is at its minimum value of v_i cosθ.
• The vertical velocity at impact has the same magnitude as v_{y0} but opposite sign: v_{yf} = −v_{y0}.

9.12 Projectile Motion Graphs and Diagrams

Four graphical representations capture different aspects of projectile motion.

The trajectory is a parabola. This can be shown by eliminating t between x = v_{x0}t and y = v_{y0}t + ½a_y t². Substituting t = x/v_{x0} into the vertical equation gives y = (v_{y0}/v_{x0})x + ½a_y(x/v_{x0})², which is a quadratic in x: the equation of a parabola.

9.13 Real-World Deviations: Air Resistance and Drag

The ideal projectile model predicts that range is maximized at 45°, complementary angles give identical ranges, and the trajectory is a perfect parabola. Real projectiles deviate from all three predictions whenever air resistance is significant.

Air resistance, or drag, acts opposite to the velocity vector and depends on speed — typically as F_{drag} ∝ v² for fast objects. This force is not constant, so the constant-acceleration equations do not apply. The effects are:

• Maximum range angle drops below 45°, typically 30–40° for balls and less for bullets.
• Complementary angles no longer give equal ranges; shallow angles outperform steep ones.
• The trajectory is no longer symmetric: the descent is steeper than the ascent.
• Impact speed is less than launch speed because energy is lost to drag.

For dense, compact objects at moderate speeds, air resistance produces corrections of order 10–20% and the ideal model is a reasonable approximation. For lightweight objects, air resistance dominates and the ideal model fails badly.

9.14 Common Mistakes in Projectile Problems

Chapter Summary

A projectile moves in two dimensions under gravity alone. The key to every projectile problem is separating motion into independent horizontal and vertical components, linked only by the shared time variable.

• Horizontal: a_x = 0; v_x = v_i cosθ = constant; x = v_x t.
• Vertical: a_y = −9.81 m/s²; v_y = v_{y0} + a_y t; y = v_{y0}t + ½a_y t².
• At the peak: v_y = 0; v_x = v_{x0}; speed = v_i cosθ.
• Time of flight for same-height launches: T = 2v_{y0}/g = 2v_i sinθ/g.
• Maximum height: H = v_{y0}²/(2g).
• Range for same-height launches: R = v_i² sin(2θ)/g; maximized at θ = 45°.
• For elevated launches, use the quadratic formula on the full vertical equation.
• The trajectory is a parabola and is symmetric about the peak for equal-height launches.
• Air resistance reduces range, lowers the optimal angle below 45°, and breaks symmetry.

Key Terms

projectile

Any object launched into the air and subsequently moving under gravity alone.

trajectory

The curved path followed by a projectile through space; a parabola under ideal conditions.

independence principle

The horizontal and vertical components of projectile motion are independent; each obeys its own kinematic equations.

v_{x0} (initial horizontal velocity)

The constant horizontal component of velocity throughout the flight: v_i cosθ.

v_{y0} (initial vertical velocity)

The initial vertical component of velocity at launch: v_i sinθ; changes during flight due to gravity.

time of flight

The total time a projectile is in the air from launch to landing.

maximum height

The greatest height above the launch point reached during the trajectory; occurs when v_y = 0.

range

The horizontal distance from the launch point to the landing point.

launch angle

The angle above the horizontal at which a projectile is launched; determines the mix of horizontal and vertical velocity.

air resistance

A drag force opposing motion through air; causes real trajectories to deviate from the ideal parabolic path.

Concept Review Questions

1. Explain in your own words why the horizontal and vertical motions of a projectile are independent.
2. A ball is launched horizontally from a cliff. Another ball is dropped from the same height at the same instant. Which lands first? Why?
3. At the peak of its trajectory, what is the vertical velocity of a projectile? What is the horizontal velocity? What is the total speed?
4. Why does a 45° launch angle maximize range when there is no air resistance?
5. Two projectiles are launched at 30° and 60° with the same speed. Compare their ranges, maximum heights, and times of flight.
6. A ball is launched horizontally from a table. Sketch the x-t, y-t, v_x-t, and v_y-t graphs for its motion from launch to landing.
7. Why does air resistance lower the optimal launch angle below 45°?
8. Why is the range formula R = v_i² sin(2θ)/g invalid for elevated launches?

Vector Decomposition Practice

Decompose each launch velocity into horizontal and vertical components. Round to three significant figures.

9. v_i = 30.0 m/s at 25.0° above horizontal.
10. v_i = 45.0 m/s at 60.0° above horizontal.
11. v_i = 12.0 m/s horizontally, 0°.
12. v_i = 18.5 m/s at 37.0° above horizontal.
13. A ball leaves a bat at 35.0 m/s at 28.0°. Find both components.
14. Given v_x = 14.0 m/s and v_y = 10.0 m/s at launch, find v_i and θ.

Projectile Motion Problems

Assume launch and landing at the same height unless stated otherwise. Use g = 9.81 m/s². Show component analysis.

15. A ball is launched at 20.0 m/s at 35.0°. Find: (a) time of flight; (b) maximum height; (c) horizontal range.
16. A soccer ball is kicked at 15.0 m/s at 40.0°. Find: (a) time to reach peak; (b) peak height; (c) range; (d) speed at peak.
17. A long jumper leaves the ground at 8.90 m/s at 23.0°. How far does she jump?
18. A ball is thrown at 25.0 m/s. What two launch angles give a range of 50.0 m?
19. At what launch angle is the range equal to the maximum height?
20. A projectile is in the air for 4.20 s and lands 120 m from the launch point. Find v_{x0} and v_{y0} at launch.

Elevated Launch Problems

21. A ball is thrown horizontally at 12.0 m/s from a 20.0 m cliff. Find: (a) time of flight; (b) range; (c) impact speed and angle.
22. A package is dropped from an airplane traveling at 90.0 m/s horizontally at 500 m altitude. Find where it lands relative to the drop point.
23. A ball is launched at 18.0 m/s at 30.0° above horizontal from a 15.0 m platform. Find: (a) time of flight; (b) horizontal range; (c) speed at impact.
24. A stone is thrown horizontally from a bridge 40.0 m above a river. It lands 55.0 m downstream. Find: (a) the launch speed; (b) the impact angle.
25. A ball is launched at 22.0 m/s at 45.0° from the edge of a 30.0 m building. Find the total time of flight and where it lands relative to the base of the building.

Conceptual Projectile Analysis

26. Sketch the trajectory of a ball launched at 45° and at 30° with the same speed. Label which goes higher and which goes farther.
27. Explain why a bullet fired horizontally and one dropped simultaneously from the same height both hit the ground at the same time.
28. A ball is thrown upward at 20.0 m/s from a moving car traveling at 15.0 m/s. Describe the trajectory as seen by: (a) the driver; (b) a person standing on the road.
29. For a given launch speed, the maximum range angle is 45°. If a quarterback must throw over a 3.0 m defensive lineman 5.0 m away, is 45° still the best angle? Explain qualitatively.

Chapter 9 Checkpoint

Before moving on, confirm that you can do each of the following.