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2.2.2 Projectile Motion

By the end of this lesson you can:

  • Explain why horizontal and vertical motion are independent
  • Calculate time of flight, range and drop for a projectile
  • Use a simulation to test your predictions
Find your course: NSW Year 11NSW Year 12: Module 5, IQ1QLD Unit 3: Gravity and motionVIC Unit 3: Motion in two dimensionsAP Physics 1: Unit 1IB: A.1 Kinematics
🎥 Watch📝 Practise🕹️ Explore🎯 Exam ready🍎 Teachers
🎥Watch
Learn the idea: a short explanation, then a check.
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Projectile motion was first systematically studied by Galileo, who designed custom apparatuses to analyze how objects move through the air. A projectile is any object moving in two dimensions—horizontally and vertically—under the influence of gravity alone.


Galileo established two fundamental principles of projectile motion:

  1. Component Motions: The horizontal component moves at a constant velocity, while the vertical component moves with constant acceleration due to gravity. This vertical acceleration causes displacement to increase proportionally with the square of time (s ∝ t2).
  2. Independence of Motion: Crucially, horizontal and vertical motions are independent of one another—what happens vertically has no effect on horizontal motion.

In essence, projectile motion consists of two simultaneous, independent motions at right angles: one moving at constant velocity, and the other undergoing uniform acceleration.

Let us explore this more closely, including the mathematical analysis required to understand projectile motion

​This video explores the concept with a number of sample situations.

It is important to not that when we study projectile motion in high school we make some assumptions, namely that of the nature of gravity and the issue of air resistance. In this vdeo I discuss that what we learn is often a simplified model

Limited time? Here is the short review version.

📝Practise
Worked examples first, then problems to try, then worksheets.
 Check your understanding
1. Ignoring air resistance, what happens to the horizontal component of a projectile's velocity while it is in the air?
2. Two identical balls are released from the same height at the exact same time: Ball A is dropped vertically from rest, while Ball B is launched horizontally. Which hits the ground first?

We are now ready to try some sample problems
Below are sample problems with videos that explain how to solve them. It is suggested you try the problems beforehand, as this actually aids understanding, even if you are unsure if you are correct.

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More problems to try
1. A bullet is fired horizontally from a rifle held 1.6 m above the ground at an initial speed of 1100 m/s. Find:
(a) The time it takes for the bullet to strike the ground.
(b) The horizontal distance traveled by the bullet.

Given: y = 1.6 m, vx = 1100 m/s, viy = 0 m/s, g = 9.8 m/s2

(a) Calculate Time (t): Using y = viyt + ½gt2

1.6 = 0 + ½(9.8)t2 → 1.6 = 4.9t2 → t2 ≈ 0.3265 → t ≈ 0.5714 s

(b) Calculate Horizontal Distance (x): x = vx × t

x = 1100 m/s × 0.5714 s ≈ 628.6 m

Answer: (a) 0.57 s, (b) 629 m
2. A major league pitcher throws a baseball horizontally at 41 m/s toward a catcher 17 m away. How much can the ball be expected to drop due to gravity during its flight?

Given: vx = 41 m/s, x = 17 m, viy = 0 m/s, g = 9.8 m/s2

Step 1 (Find Time): t = x / vx = 17 / 41 ≈ 0.4146 s

Step 2 (Calculate Vertical Drop): y = ½gt2

y = ½(9.8)(0.4146)2 ≈ 0.842 m

Answer: 0.84 m
3. A horizontal rifle is aimed directly at the center of a bull's eye. The muzzle speed of the bullet is 670 m/s. If the bullet strikes the target 2.5 cm below the center, what is the horizontal distance between the end of the rifle and the bull's eye?

Given: vx = 670 m/s, y = 2.5 cm = 0.025 m, g = 9.8 m/s2

Step 1 (Find Time from Vertical Drop): y = ½gt2

0.025 = 4.9t2 → t2 ≈ 0.005102 → t ≈ 0.07143 s

Step 2 (Calculate Distance): x = vx × t

x = 670 m/s × 0.07143 s ≈ 47.86 m

Answer: 47.9 m
4. A jet fighter is traveling horizontally at a speed of 111 m/s at an altitude of 300 m when an outboard fuel tank is released.
(a) How much time elapses before the tank hits the ground?
(b) What is the velocity (magnitude and direction) of the tank just before impact?

Given: vx = 111 m/s, y = 300 m, viy = 0 m/s, g = 9.8 m/s2

(a) Calculate Time (t): y = ½gt2 → 300 = 4.9t2 → t ≈ 7.82 s

(b) Calculate Final Velocity:

vfy = gt = 9.8 × 7.82 ≈ 76.68 m/s (downward)

Magnitude: v = √(vx2 + vfy2) = √(1112 + 76.682) ≈ 134.9 m/s

Direction: θ = tan−1(vfy / vx) = tan−1(76.68 / 111) ≈ 34.6° below the horizontal

Answer: (a) 7.82 s, (b) 134.9 m/s at 34.6° below horizontal
🕹️Explore
Test your predictions in a simulation.

Lets Play

  1. Explore the Interface: Start with the Intro tab to familiarize yourself with the controls.
  2. Experiment: Fire a few test rounds while adjusting initial parameters like launch height, speed, and angle.
  3. Observe Vectors: Toggle the velocity and acceleration vector displays to visualize how the projectile behaves in flight.
  4. Calculate: Set your own initial values, then calculate the expected range manually before firing.
  5. Verify: Move the target to your calculated range and test your prediction!


I have got these from a variety of sources, some my own, some from elsewhere. Lots of practice here.

🎯Exam ready
Check you can do it under exam conditions.
How Well do you know Projectiles?
Test your understanding of projectiles. So do the Quiz and try to get full marks
Then check your understanding if necessary with the video