Unit 03 - Grade 11-12 Physics

Projectile Motion

Connect vectors and kinematics to study objects launched through the air. Learn horizontal and vertical components, time of flight, range, maximum height, launch angle, and motion graphs.

Lesson roadmap

What Students Should Master in This Unit

Projectile motion is two-dimensional motion under gravity. The powerful idea is that horizontal and vertical motion happen at the same time, but they can be analyzed separately.

Separate components

Break launch velocity into horizontal and vertical parts using sine and cosine.

Use time correctly

Recognize that horizontal and vertical motion share the same time of flight.

Solve flight paths

Find range, height, velocity, landing time, and graph features for projectile motion.

Quick navigation

Jump to a Topic

Core Idea Assumptions Vector Components Horizontal Motion Vertical Motion Projectile Equations Horizontal Launch Angled Launch Range and Height Projectile Graphs Worked Examples Practice Problems
Big picture

1. What Is Projectile Motion?

A projectile is an object that moves through the air while gravity is the main influence on its motion. After launch, the object continues moving horizontally while gravity pulls it vertically downward.

Projectile motion overview Projectile motion combines x-motion and y-motion vx stays constant ay = -g launch landing
The motion is easier when students solve the horizontal and vertical parts separately, then connect them with the same time.
Horizontal motion ax = 0 Horizontal velocity stays constant if air resistance is ignored.
Vertical motion ay = -g = -9.8 m/s2 Use this if upward is positive.
Shared time tx = ty Both components describe the same object during the same flight.
Main strategy: Split every projectile problem into an x-table and a y-table. Use the same time value in both directions.
Model limits

2. Assumptions in Basic Projectile Motion

Grade 11-12 projectile problems usually use an idealized model. The model is powerful, but students must know what it assumes.

  • Air resistance is ignored unless the question says otherwise.
  • Gravity is constant near Earth's surface: g = 9.8 m/s2.
  • The horizontal acceleration is zero.
  • The vertical acceleration is downward.
  • Horizontal and vertical motion are independent except for sharing the same time.
  • The projectile's path is a parabola when air resistance is ignored.
Ideal projectile assumptions Ideal model: gravity acts downward, air resistance is ignored g = 9.8 m/s2 air drag not used included in basic problems usually ignored unless stated
The standard Grade 11-12 model keeps gravity constant and ignores drag, so calculations stay focused on components and time.
Common mistake: Do not use the total launch speed in a horizontal equation unless the launch is purely horizontal. For angled launches, use vx horizontally and vy vertically.
Launch velocity

3. Breaking Velocity into Components

If a projectile is launched at an angle, its velocity has a horizontal component and a vertical component. These components act like separate starting velocities for the x and y directions.

Velocity components Resolve launch velocity into perpendicular components v0 vx = v0cos(θ) vy0 = v0sin(θ) θ
Use cosine for the side adjacent to the launch angle and sine for the side opposite the launch angle.
Horizontal component vx = v0 cos(θ) Use when the angle is measured above the horizontal.
Vertical component vy0 = v0 sin(θ) Positive if launched upward and upward is chosen positive.
Speed from components v = √(vx2 + vy2) Use for total velocity at any instant.
Direction from components θ = tan-1(vy / vx) Check whether the projectile is moving up or down.
Horizontal velocity vx = constant Only changes if air resistance is included.
Vertical velocity vy = vy0 - gt Use upward positive.
x-direction

4. Horizontal Motion

In basic projectile motion, no horizontal force is included after launch, so horizontal acceleration is zero. This means horizontal velocity stays constant.

Horizontal motion Horizontal velocity is constant in the ideal model equal Δx equal Δx equal Δx ax = 0 Δx = vxt
Equal horizontal spacing in equal time intervals means the horizontal speed is not changing.
Horizontal acceleration ax = 0 No horizontal acceleration in the ideal model.
Horizontal displacement Δx = vxt This is often used to find range.
Horizontal velocity vx = Δx / t Useful for horizontally launched projectiles.

Meaning

If two objects fall from the same height at the same time, one dropped and one launched horizontally, they hit the ground at the same time if air resistance is ignored. The horizontal velocity does not affect falling time.

y-direction

5. Vertical Motion

The vertical direction behaves like free fall. Gravity changes the vertical velocity by 9.8 m/s every second, downward.

Vertical motion Gravity changes vertical velocity throughout flight vy upward ay = -g vy = 0 at top vy downward
At the top, only vertical velocity is zero. Acceleration is still downward because gravity is still acting.
Vertical acceleration ay = -9.8 m/s2 Use if upward is positive.
Vertical velocity vy = vy0 + ayt Use to find velocity at a time.
Vertical displacement Δy = vy0t + (1/2)ayt2 Use to find height or time.
No-time vertical equation vy2 = vy02 + 2ayΔy Helpful for max height.
At the top vy = 0 Only the vertical velocity is zero at the peak.
Acceleration at the top ay = -9.8 m/s2 Gravity still acts at the peak.
Formula toolkit

6. Projectile Motion Equation Set

Most projectile problems are solved by writing separate x and y equations. Horizontal equations are constant-velocity equations. Vertical equations are constant-acceleration equations.

Projectile equation setup Set up separate x and y columns before solving Horizontal x-motion Vertical y-motion ax = 0 vx = constant Δx = vxt ay = -g vy = vy0 - gt Δy = vy0t - (1/2)gt2 same t
The safest method is to list knowns separately, solve vertical motion for time when needed, then return to horizontal motion for range.
Direction Acceleration Velocity Displacement Best Use
Horizontal ax = 0 vx = constant Δx = vxt Range, horizontal speed, flight time.
Vertical ay = -g vy = vy0 - gt Δy = vy0t - (1/2)gt2 Height, landing time, vertical velocity.

Problem-Solving Method

  1. Draw the path and choose positive directions.
  2. Split the initial velocity into vx and vy0 if needed.
  3. Make an x-table and y-table of knowns.
  4. Use vertical motion to find time if height information is given.
  5. Use horizontal motion to find range or horizontal velocity.
  6. Combine velocity components if the final speed or direction is asked.
Special case

7. Horizontal Launch Problems

In a horizontal launch, the projectile starts with horizontal velocity but no initial vertical velocity. This often describes a ball rolling off a table or a package moving off a ledge.

Horizontal launch A horizontal launch starts with vy0 = 0 vx g Δx = vxt table height gives time
For a horizontal launch, solve the fall time from vertical motion first, then use that same time for horizontal range.
Initial vertical velocity vy0 = 0 For horizontal launch only.
Falling height Δy = -(1/2)gt2 Use upward positive and downward displacement negative.
Time from height t = √(2h / g) Works when launched horizontally from height h.
Range Δx = vxt Find time from vertical motion first.
Final vertical velocity vy = -gt Use upward positive.
Impact speed v = √(vx2 + vy2) Combine components at impact.
Launch angle

8. Angled Launch Problems

An angled launch has both horizontal and vertical initial velocity. The horizontal component stays constant, while the vertical component changes because of gravity.

Angled launch Same launch and landing height gives useful symmetry v0 top: vy = 0 range R time up time down
When the projectile lands at the same height it was launched, the upward and downward parts mirror each other in time.

Symmetry for Same Launch and Landing Height

  • Time going up equals time coming down.
  • At the top, vy = 0.
  • The speed just before landing equals the launch speed if air resistance is ignored.
  • The vertical landing velocity is the negative of the initial vertical velocity.
  • Complementary angles, such as 30 degrees and 60 degrees, give the same range if launch speed is the same and landing height is the same.
Time to top ttop = vy0 / g Use when landing height equals launch height or for peak timing.
Total time, same height ttotal = 2vy0 / g Only when launch and landing heights are equal.
Horizontal range R = vxttotal This is the safest range method.
Peak and landing

9. Range, Maximum Height, and Time of Flight

Some formulas are shortcuts that work only under specific conditions. Use them carefully and know their limits.

Range and maximum height Range, maximum height, and flight time describe the full path H R maximum height t = 0 ttotal
Use shortcut formulas only when the launch and landing heights are the same; otherwise find time from vertical motion first.
Maximum height above launch point H = vy02 / (2g) Works when air resistance is ignored.
Time to peak ttop = vy0 / g Vertical velocity is zero at the top.
Total time, same height t = 2vy0 / g Only for same launch and landing height.
Range, same height R = (v02 sin(2θ)) / g Shortcut for same launch and landing height only.
Maximum range angle θ = 45 degrees Only on level ground without air resistance.
General range method R = vxt Works after the correct flight time is found.
Professional habit: If launch and landing heights are different, avoid the shortcut range formula. Solve vertical motion for time first, then use horizontal motion.
Visual motion

10. Projectile Motion Graphs

Projectile graphs help students see why the motion separates into independent horizontal and vertical parts.

Projectile graph shapes Graph shapes reveal constant x-motion and accelerated y-motion x vs. t: line y vs. t: parabola vy vs. t: negative slope
Horizontal position grows steadily, vertical position curves, and vertical velocity changes linearly because acceleration is constant.
Graph Shape Meaning
x vs. tStraight lineHorizontal velocity is constant.
vx vs. tHorizontal lineHorizontal velocity does not change.
ax vs. tLine at zeroNo horizontal acceleration.
y vs. tParabolaVertical position changes under constant acceleration.
vy vs. tStraight line with negative slopeVertical velocity decreases by 9.8 m/s every second.
ay vs. tHorizontal line at -9.8 m/s2Vertical acceleration is constant downward.

Path Shape

The actual path through space is parabolic because horizontal motion is constant while vertical motion accelerates downward.

Investigation skills

11. Projectile Motion Lab Skills

Projectile labs usually connect video analysis, launch speed, launch angle, range, and landing height. Students should be able to compare measured results to calculated predictions.

Projectile motion lab setup Good labs compare measured range with calculated range launcher video camera target measured R
Professional lab work records launch angle, launch speed, height, measured range, uncertainty, and percent error.

Common Labs

  • Roll a ball off a table and predict where it lands.
  • Launch a ball at different angles and measure range.
  • Use video analysis to track x-position and y-position frame by frame.
  • Compare measured range to calculated range and find percent error.
  • Investigate which launch angle produces maximum range on level ground.

Lab Report Questions

  • What assumptions were made about air resistance?
  • How was launch speed measured?
  • How was time of flight determined?
  • Which measurement created the greatest uncertainty?
  • Was the calculated landing position inside the experimental uncertainty range?
Worked examples

12. Worked Examples

Example 1: Components of launch velocity

A ball is launched at 20.0 m/s at 30.0 degrees above the horizontal. Find vx and vy0.

vx = v0cos(θ) = 20.0cos(30.0) = 17.3 m/s.

vy0 = v0sin(θ) = 20.0sin(30.0) = 10.0 m/s.

Example 2: Horizontal launch from a table

A ball rolls off a 1.25 m high table with horizontal speed 3.00 m/s. Find time to hit the floor and horizontal range.

Vertical: h = (1/2)gt2, so t = √(2h/g).

t = √[2(1.25)/9.8] = 0.505 s.

Horizontal: Δx = vxt = (3.00)(0.505) = 1.52 m.

Example 3: Time to maximum height

A projectile has initial vertical velocity 14.7 m/s upward. Find time to reach the highest point.

At the top, vy = 0.

vy = vy0 - gt, so 0 = 14.7 - 9.8t.

t = 1.50 s.

Example 4: Maximum height

A projectile is launched with vy0 = 18.0 m/s. Find maximum height above the launch point.

H = vy02 / (2g).

H = (18.0)2 / [2(9.8)] = 16.5 m.

Example 5: Range on level ground

A ball is launched at 25.0 m/s at 40.0 degrees and lands at the same height. Find range.

vx = 25.0cos(40.0) = 19.2 m/s.

vy0 = 25.0sin(40.0) = 16.1 m/s.

Total time = 2vy0/g = 2(16.1)/9.8 = 3.28 s.

Range = vxt = (19.2)(3.28) = 63.0 m.

Example 6: Impact velocity

A projectile has vx = 12.0 m/s and vy = -16.0 m/s just before impact. Find impact speed and direction below horizontal.

v = √(12.02 + 16.02) = 20.0 m/s.

θ = tan-1(16.0/12.0) = 53.1 degrees below the horizontal.

Independent practice

13. Practice Problems

Try each problem first. Then open the answer check and compare your setup, signs, and units.

1. A projectile is launched at 10.0 m/s horizontally from a 20.0 m high cliff. Find the time of flight.

Answer

t = √(2h/g) = √[2(20.0)/9.8] = 2.02 s.

2. Using problem 1, find the horizontal range.

Answer

Δx = vxt = (10.0)(2.02) = 20.2 m.

3. A ball is launched at 30.0 m/s at 60.0 degrees. Find vx.

Answer

vx = 30.0cos(60.0) = 15.0 m/s.

4. A ball is launched at 30.0 m/s at 60.0 degrees. Find vy0.

Answer

vy0 = 30.0sin(60.0) = 26.0 m/s.

5. A projectile has vy0 = 19.6 m/s. Find time to maximum height.

Answer

ttop = vy0/g = 19.6/9.8 = 2.00 s.

6. For problem 5, find total flight time if it lands at the launch height.

Answer

Total time = 2ttop = 4.00 s.

7. A projectile has vx = 18.0 m/s and stays in the air for 3.50 s. Find range.

Answer

R = vxt = (18.0)(3.50) = 63.0 m.

8. A ball is launched upward with vy0 = 12.0 m/s. Find maximum height.

Answer

H = vy02/(2g) = 12.02 / 19.6 = 7.35 m.

9. At the top of a projectile's path, what is vy?

Answer

vy = 0.

10. At the top of a projectile's path, what is acceleration?

Answer

Acceleration is still downward, 9.8 m/s2. If upward is positive, ay = -9.8 m/s2.

11. A ball is launched at 22.0 m/s and 35.0 degrees. Find horizontal velocity.

Answer

vx = 22.0cos(35.0) = 18.0 m/s.

12. A ball is launched at 22.0 m/s and 35.0 degrees. Find initial vertical velocity.

Answer

vy0 = 22.0sin(35.0) = 12.6 m/s.

13. A projectile lands at the same height it was launched from. Which angle gives the greatest range if speed is fixed?

Answer

45 degrees, assuming no air resistance.

14. A projectile lands at the same height it was launched from. Which two angles have the same range as 30 degrees?

Answer

30 degrees and 60 degrees are complementary and produce the same range for the same launch speed.

15. A horizontal launch has vx = 6.0 m/s and impact vy = -8.0 m/s. Find impact speed.

Answer

v = √(6.02 + 8.02) = 10.0 m/s.

16. A velocity-time graph for vy has slope -9.8 m/s2. What does that represent?

Answer

The vertical acceleration due to gravity.

Final review

14. What to Know Before Moving On

  • Projectile motion separates into horizontal and vertical components.
  • Horizontal velocity is constant when air resistance is ignored.
  • Vertical acceleration is 9.8 m/s2 downward.
  • Horizontal and vertical motion share the same time.
  • At the top of the path, vertical velocity is zero but acceleration is not zero.
  • Use vx = v0cos(θ) and vy0 = v0sin(θ) for angled launches.
  • For different launch and landing heights, solve vertical motion for time before finding range.
  • The ideal path of a projectile is a parabola.
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