📘 CodingMarble Learn

Projectile Motion and Uniform Circular Motion (Class 11)

In a plane, r = r₀ + v₀t + ½at² and v = v₀ + at, applied separately along x and y. A projectile has constant horizontal velocity u cos θ and a vertical velocity that changes by g each second, so its path is a parabola: y = x tan θ − gx²/(2u²cos²θ). T = 2u sin θ/g, H = u² sin²θ/2g, R = u² sin 2θ/g (maximum at 45°). In uniform circular motion speed is constant but the velocity turns, giving a centripetal acceleration a = v²/r = ω²r towards the centre.

🎬 Step-by-step story

  1. In a plane, uniform velocity gives equal steps along a straight line: r = r₀ + vt. Add a steady acceleration and the velocity arrow turns, so the path bends.
  2. A thrown ball: its sideways velocity never changes. Its upward velocity drops by 10 m/s every second, becomes zero at the top, then points down. The two parts do not disturb each other.
  3. Put x = (u cos θ)t into y = (u sin θ)t − ½gt² and remove t. You get y = bx − cx²: the path is a parabola.
  4. From the top point we get time of flight, maximum height and range. 30° and 60° land at the same spot. 45° goes the farthest.
  5. Circular motion: the speed stays the same, but the direction keeps turning. The velocity is along the tangent; the acceleration points to the centre, a = v²/r.
  6. Your turn: choose the launch speed and the angle. Watch the path and read T, H and R below.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why does the path bend when acceleration is added?

Acceleration changes the velocity arrow a little each second. When it is not along the velocity, the arrow turns, and the path turns with it.

Why does gravity not slow down the horizontal motion?

Gravity pulls straight down, so it has no sideways part. In the 3D the blue arrow keeps the same length all the way.

Is the velocity zero at the top?

No. Only the vertical part is zero. The horizontal part keeps the ball moving sideways.

Why is the path a parabola and not a circle?

x grows like t but the drop grows like t². Removing t gives y = bx − cx², which is the equation of a parabola.

Why do 30° and 60° give the same range?

Range depends on sin 2θ, and sin 60° = sin 120°. The 60° throw goes higher but moves sideways slower; the two effects balance.

If the speed is constant in a circle, why is there an acceleration?

The velocity keeps turning. A change of direction is a change of velocity, and that change always points to the centre.

Uniform velocity and uniform acceleration in a plane

In a plane, position is r = x î + y ĵ. Velocity v = dr/dt = vx î + vy ĵ, and acceleration a = dv/dt.

Uniform velocity: v constant → r = r₀ + vt, a straight path.

Uniform acceleration: v = v₀ + at and r = r₀ + v₀t + ½at². These are really two straight-line problems: x = x₀ + v₀ₓt + ½aₓt² and y = y₀ + v₀ᵧt + ½aᵧt². The x and y motions are independent.

The direction of v is along the tangent to the path at every point.

Projectile motion

A projectile is any object thrown into the air that then moves only under gravity (air resistance ignored). Launch speed u at angle θ above the horizontal:

Equation of the path (trajectory)

t = x/(u cos θ) → y = x tan θ − g x²/(2u² cos²θ), of the form y = bx − cx², a parabola.

Time of maximum height, time of flight, height and range

At the top vy = 0 → tₘ = u sin θ/g. Time of flight T = 2u sin θ/g. Maximum height H = u² sin²θ/2g. Horizontal range R = u² sin 2θ/g, largest at θ = 45° (Rₘₐₓ = u²/g). Angles θ and 90° − θ give the same range.

Horizontal launch from a height h

uy = 0, so t = √(2h/g) and range = u√(2h/g).

Uniform circular motion

An object going round a circle at constant speed is in uniform circular motion. Its velocity keeps changing direction, so it is accelerating.

Angular speed ω = Δθ/Δt (rad/s); v = ωr; time period T = 2πr/v = 2π/ω; frequency ν = 1/T, so ω = 2πν.

Centripetal acceleration

In a short time Δt the velocity vector turns by Δθ. The change |Δv| ≈ vΔθ and it points towards the centre. So a = |Δv|/Δt = vω = v²/r = ω²r, always towards the centre. Its size is constant, but its direction keeps changing, so it is not a constant vector.

Key formulas and definitions

Worked examples

1. A particle starts at the origin with v₀ = 4î m/s and a = 2ĵ m/s². Find its position and velocity at t = 3 s.

Step 1: x = 4 × 3 = 12 m (no x-acceleration). Step 2: y = ½ × 2 × 9 = 9 m. Step 3: v = 4î + 6ĵ m/s, speed = √(16 + 36) ≈ 7.2 m/s.

2. A ball is thrown at 20 m/s at 30° above the ground (g = 10 m/s²). Find its time of flight.

Step 1: uy = 20 sin 30° = 10 m/s. Step 2: T = 2uy/g = 2 × 10/10 = 2 s.

3. For the same throw (20 m/s at 30°), find the maximum height and the range.

Step 1: H = uy²/2g = 100/20 = 5 m. Step 2: ux = 20 cos 30° ≈ 17.32 m/s. Step 3: R = ux × T = 17.32 × 2 ≈ 34.6 m (or u² sin 60°/g = 400 × 0.866/10).

4. What is the greatest distance a ball thrown at 30 m/s can cover on level ground (g = 10 m/s²)?

Step 1: range is maximum at 45°. Step 2: Rₘₐₓ = u²/g = 900/10 = 90 m.

5. A stone is thrown horizontally at 15 m/s from a 20 m high roof (g = 10 m/s²). When and where does it land?

Step 1: uy = 0 → 20 = ½ × 10 × t² → t = 2 s. Step 2: x = 15 × 2 = 30 m from the foot of the building. Step 3: at landing vy = 20 m/s, so speed = √(225 + 400) = 25 m/s.

6. Show that throwing at 25° and 65° with the same speed gives the same range.

Step 1: R ∝ sin 2θ. Step 2: sin 50° and sin 130° are equal because sin(180° − x) = sin x. Step 3: so R(25°) = R(65°); the 65° throw goes higher and stays up longer.

7. A stone tied to a 0.5 m string goes round 2 times every second. Find its speed and centripetal acceleration.

Step 1: ω = 2πν = 4π ≈ 12.57 rad/s. Step 2: v = ωr ≈ 6.28 m/s. Step 3: a = ω²r = 16π² × 0.5 ≈ 79 m/s², towards the centre.

8. A car goes round a 50 m radius turn at 36 km/h. Find its centripetal acceleration.

Step 1: 36 km/h = 10 m/s. Step 2: a = v²/r = 100/50 = 2 m/s². Step 3: direction: towards the centre of the turn.

Common mistakes

Practice quiz

1. At the highest point of a projectile's path, the acceleration is:
2. The range is the same for angles:
3. The path of a projectile is a:
4. In uniform circular motion, the acceleration points:
5. Doubling the launch speed makes the range:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is projectile motion in Class 11?

Motion of a body thrown into the air and moving under gravity alone. It combines constant horizontal velocity with uniformly accelerated vertical motion.

Why is the range maximum at 45°?

R = u² sin 2θ/g and sin 2θ has its largest value 1 when 2θ = 90°, that is θ = 45°.

What is centripetal acceleration?

The acceleration of a body moving in a circle, directed towards the centre, with size v²/r = ω²r.

Where this is taught

Spain1º BachilleratoKinematics
Ukraine10 класMechanics
Ukraine10 класMechanics
CBSE (India)Class 11Kinematics
England (GCSE, A level)Year 13Q Kinematics (2D and projectiles)
USA (Common Core, NGSS, AP)Grade 12Kinematics
Japan高校2年Various motions
South Korea고등학교 2학년Space-time and motion
South Korea고등학교 3학년Mechanical interactions
Russia9 классMechanical phenomena
Russia10 классMechanics: kinematics
China高一Compulsory 2 Ch.5 Projectile motion

Learn first

Learn next

Related lessons

All Physics lessons