The projectile motion formula describes an object launched at angle θ with initial speed u, moving under gravity alone. The three core results are: time of flight T = 2u sinθ / g, maximum height H = u² sin²θ / 2g, and horizontal range R = u² sin2θ / g. These come from splitting the motion into two independent parts, constant horizontal velocity and uniformly accelerated vertical motion, then applying the standard equations of motion to each direction separately.
What Is Projectile Motion?
Projectile motion physics describes any object thrown into the air that moves under the influence of gravity alone, once it leaves the hand, bat, or launcher.
A thrown ball, a kicked football, and a bullet fired at an angle are all projectiles, and the curved path each one traces is called its trajectory.
Seeing this split into two independent motions demonstrated visually in a narrated video walkthrough tends to make the idea click faster than reading the definition alone.
The defining feature of projectile motion is that it splits cleanly into two independent motions happening at the same time. Horizontally, there is no force acting on the object (ignoring air resistance), so the horizontal velocity stays constant throughout the flight. Vertically, gravity constantly pulls the object downward, producing uniform acceleration.
According to NASA’s Glenn Research Center, the flight path of a ballistic object depends only on its initial velocity and gravitational acceleration, and includes no information at all about the object’s size, shape, or mass, which is why a cricket ball and a shot put launched at the same angle and speed follow identical paths.
Projectile Motion Formula
The core projectile motion equations below assume the object is launched from ground level with initial speed u at angle θ above the horizontal, and that air resistance is ignored.
| Quantity | Formula | What It Tells You |
|---|---|---|
|
Time of Flight (T) |
T = 2u sinθ / g |
Total time the projectile stays in the air |
|
Maximum Height (H) |
H = u² sin²θ / 2g |
Highest vertical point the projectile reaches |
|
Horizontal Range (R) |
R = u² sin2θ / g |
Total horizontal distance covered |
|
Equation of Trajectory |
y = x tanθ − gx² / (2u² cos²θ) |
The parabolic path traced by the projectile |
|
Maximum Range |
R_max = u² / g, at θ = 45° |
Greatest possible range for a given speed |
Here, u is the initial launch speed, θ is the angle of projection measured from the horizontal, and g is the acceleration due to gravity, taken as 9.8 m/s² near Earth’s surface.
Projectile Motion Derivation
Here is the projectile motion derivation for the three main formulas, starting from the basic equations of motion applied separately to the horizontal and vertical directions.
Setting up the components
At launch, the initial velocity u splits into two components:
- Horizontal component: uₓ = u cosθ
- Vertical component: uᵧ = u sinθ
The horizontal component stays constant throughout the flight, since there is no horizontal force. The vertical component decreases as the object rises, reaches zero at the peak, then increases again as the object falls.
Deriving time of flight
The projectile lands when its vertical displacement returns to zero. Using s = ut + ½at² for the vertical direction, with initial vertical velocity u sinθ and acceleration −g:
0 = (u sinθ)T − ½g T²
Solving for T (excluding the trivial solution T = 0) gives:
T = 2u sinθ / g
Deriving maximum height
At the highest point, the vertical velocity becomes zero. Using v² = u² − 2as for the vertical direction:
0 = (u sinθ)² − 2gH
Solving for H:
H = u² sin²θ / 2g
Deriving horizontal range
Since horizontal velocity is constant, range is simply horizontal velocity multiplied by total time of flight:
R = (u cosθ) × T = (u cosθ) × (2u sinθ / g)
Using the identity 2 sinθ cosθ = sin2θ, this simplifies to:
R = u² sin2θ / g
Projectile Motion Formula Class 11: Where This Fits In
Projectile motion class 11 sits inside Chapter 3, Motion in a Plane, part of the official NCERT physics curriculum. The chapter builds up from scalars and vectors, through vector addition, to projectile motion and uniform circular motion, using the same vector components covered here.
Since these formulas assume no air resistance and a launch from ground level, exam questions often adjust one of those assumptions (launching from a height, or asking for velocity at a specific point), so it helps to understand the derivation, not just memorize the final formulas. A worked video example of one of these modified problems is often the fastest way to see how the adjustment changes the setup.
Projectile Motion Examples
Example 1: Finding time of flight and range
A ball is thrown with an initial speed of 20 m/s at an angle of 30° above the horizontal. Find the time of flight and horizontal range. (Take g = 10 m/s².)
T = 2u sinθ / g = 2(20)(sin30°) / 10 = 2(20)(0.5) / 10 = 2 seconds
R = u² sin2θ / g = (20)² sin60° / 10 = 400(0.866) / 10 = 34.6 meters
Example 2: Finding maximum height
A cricketer throws a ball at 25 m/s at an angle of 60° to the horizontal. Find the maximum height reached. (Take g = 10 m/s².)
H = u² sin²θ / 2g = (25)² (sin60°)² / 2(10) = 625(0.75) / 20 = 23.4 meters
Example 3: Finding the launch angle for maximum range
At what angle should a javelin be thrown to achieve maximum horizontal range?
Since R = u² sin2θ / g, range is maximum when sin2θ = 1, meaning 2θ = 90°, so θ = 45°. This is why 45° is the optimal launch angle for maximum range on level ground, a result that shows up constantly in projectile motion questions on optimization.
Projectile Motion Problems and Numericals
Try these projectile motion problems, then check your answers against the solutions below. Use g = 9.8 m/s² unless stated otherwise.
Problem 1: A stone is projected at 15 m/s at an angle of 45°. Find its time of flight.
Solution: T = 2(15)(sin45°) / 9.8 = 2(15)(0.707) / 9.8 ≈ 2.16 seconds.
Problem 2: A football is kicked at 18 m/s at 37° to the horizontal. Find the maximum height reached.
Solution: H = (18)² (sin37°)² / (2 × 9.8) = 324(0.362) / 19.6 ≈ 5.99 meters.
Problem 3: A projectile has a range of 40 m when launched at 30° with initial speed u. Find u.
Solution: R = u² sin60° / g, so 40 = u²(0.866) / 9.8, giving u² = 452.7, so u ≈ 21.3 m/s.
Checklist before you submit an answer:
- Did you confirm g is the value given in the question (9.8 or 10 m/s²)?
- Did you use sinθ for height-related formulas and sin2θ for range?
- Did you keep angle units consistent (degrees vs. radians) throughout the calculation?
- Did you check your final answer’s units (seconds for time, meters for height and range)?
Working through these projectile motion numericals with a step-by-step video explanation helps confirm not just the final number, but which formula applies and why, before the next numerical shows up in an exam.
How Think10x.ai Helps With Projectile Motion
Projectile motion questions look simple on the surface, but small mix-ups, using sinθ where cosθ belongs, forgetting to square a term, or mixing up which formula uses θ and which uses 2θ, produce wrong answers even when the concept is understood.
Think10x.ai turns a projectile motion question, typed, spoken, or photographed from a worksheet, into a narrated video that walks through the vector decomposition and each formula individually, and students can pause to ask why a particular substitution happens.
In partnership with Vidyamandir Classes (VMC), this approach helped clear over 80% of student doubts independently across more than 8,500 students, exactly the kind of formula-selection confusion that mechanics chapters like this one tend to create.
Teachers preparing projectile motion questions for practice sets can generate full walkthroughs using the platform’s AI math solver, and students revising independently get the same explanation depth through the Think10x.ai homepage.
Frequently Asked Questions
Because gravity acts only in the vertical direction, and air resistance is ignored in the standard formulas. With no horizontal force acting on the object, its horizontal velocity cannot change during flight.
A parabola. This comes directly from the trajectory equation y = x tanθ − gx² / (2u² cos²θ), which is a quadratic relationship between x and y.
45°. Since range depends on sin2θ, and sine reaches its maximum value of 1 at 90°, the range is greatest when 2θ = 90°, meaning θ = 45°.
Yes. These are complementary angles that add up to 90°, and sin2θ gives the same value for both, so a projectile launched at 30° or 60° with the same initial speed covers the same horizontal range, though the time of flight and maximum height differ.
No. The standard formulas assume air resistance is negligible. Real projectiles like a badminton shuttlecock or a golf ball experience drag, which shortens the range and changes the shape of the trajectory from a true parabola.