Motion in a Plane: Projectile and Circular Motion Formulas with JEE PYQs

Quick Answer: The core projectile motion formulas for ground-to-ground projection at angle θ with speed u are: Time of flight T = 2u sinθ / g, Maximum height H = u² sin²θ / 2g, and Range R = u² sin2θ / g (maximum at θ = 45°). For uniform circular motion, centripetal acceleration is a = v²/r = ω²r, directed toward the centre, with centripetal force F = mv²/r and linear speed v = ωr.

Motion in a Plane: Basics You Must Know

Before tackling two-dimensional motion, you must be comfortable distinguishing scalars from vectors. A scalar (mass, speed, time, energy) has only magnitude. A vector (displacement, velocity, acceleration, force) has magnitude and direction, and obeys vector addition rules — the triangle or parallelogram law.

The single most powerful idea in planar motion is this:

Motion in a plane is simply two independent one-dimensional motions happening at the same time. The x-axis does not know what the y-axis is doing. Gravity pulls only downward; it has zero effect on horizontal motion (when air resistance is ignored). Time is the only “bridge” linking the two axes — the object spends the same t seconds moving horizontally as it does vertically.

To work in 2D, we resolve vectors into components. A vector A making angle θ with the x-axis decomposes as:

• A_x = A cosθ (component along x)
• A_y = A sinθ (component along y)
• Magnitude: A = √(A_x² + A_y²)

Projectile Motion: Horizontal and Vertical Decomposition

A projectile is any object given an initial velocity and then moving under gravity alone. Consider a body launched from the ground with initial speed u at angle θ to the horizontal.

Step 1 — Resolve the initial velocity:

• Horizontal component: u_x = u cosθ
• Vertical component: u_y = u sinθ

Step 2 — Apply 1D equations to each axis independently:

Because a_x = 0, the horizontal velocity never changes. The vertical velocity decreases by g every second on the way up, becomes zero at the highest point, and increases on the way down. The trajectory traced out is a parabola: y = x tanθ − (g x²)/(2u² cos²θ).

Time of Flight, Maximum Height, and Horizontal Range Formulas

All three classic formulas come from the vertical and horizontal equations above.

Time of Flight (T): The projectile returns to the ground when y = 0:

0 = (u sinθ)T − ½gT²  ⇒  T = 2u sinθ / g

Maximum Height (H): At the top, v_y = 0. Using v² = u² − 2as:

0 = (u sinθ)² − 2gH  ⇒  H = u² sin²θ / 2g

Horizontal Range (R): Distance covered horizontally in time T:

R = (u cosθ)(T) = u² sin2θ / g

Since the maximum value of sin2θ is 1 (at 2θ = 90°, i.e., θ = 45°), the range is maximum at θ = 45°, and R_max = u²/g.

Special Cases: Horizontal Projection and Equal Ranges

Horizontal projection from a height h (e.g., a stone thrown horizontally from a cliff, or a bomb dropped from a plane): Here θ = 0, so u_x = u, u_y = 0.

• Time to fall: t = √(2h/g) — independent of u!
• Horizontal distance covered: R = u√(2h/g)
• Vertical velocity on impact: v_y = √(2gh)

Equal ranges at complementary angles: Since R = u² sin2θ/g and sin2θ = sin(180° − 2θ) = sin2(90° − θ), projection at angle θ and at angle (90° − θ) gives the same range. Example: 30° and 60°. But note — the times of flight and maximum heights are different (the larger angle flies higher and longer in time).

Projectile Motion Formula Sheet (One-Page Revision)

Uniform Circular Motion: Centripetal Acceleration and Force

In uniform circular motion, a body moves in a circle of radius r with constant speed v. The speed (magnitude) is constant, but the direction of the velocity changes continuously — so there must be an acceleration.

That acceleration always points toward the centre of the circle and is called centripetal acceleration:

a_c = v²/r = ω²r

The corresponding centripetal force is F = mv²/r = mω²r, directed toward the centre. It is provided by some real force — tension, friction, gravity, or the normal reaction — never “by itself”.

Angular variables you must know:

• Angular displacement θ (in radians)
• Angular velocity ω = dθ/dt = 2π/T = 2πf (rad/s)
• Angular acceleration α = dω/dt (rad/s²)
• Linear speed v = ωr
• In one full revolution: displacement = 0, average velocity = 0, average speed = 2πr/T

Circular Motion Formula Sheet and Relations

For deeper practice aligned with the official syllabus, refer to the JEE Main official portal and NCERT Physics Class 11, Chapter 3 (Motion in a Plane).

Solved JEE PYQ 1: Projectile (Time of Flight / Range)

Question (JEE Main pattern): A ball is thrown with speed 20 m/s at 30° above the horizontal from level ground. Find its time of flight and horizontal range. (Take g = 10 m/s²)

Step 1 — Decompose the velocity:

• u_x = 20 cos30° = 20 × (√3/2) ≈ 17.32 m/s
• u_y = 20 sin30° = 20 × 0.5 = 10 m/s

Step 2 — Time of flight:

T = 2u_y/g = 2(10)/10 = 2 s

Step 3 — Range:

R = u_x × T = 17.32 × 2 ≈ 34.64 m

Verification with the direct formula: R = u² sin2θ/g = (400 × sin60°)/10 = 400 × 0.866/10 ≈ 34.64 m ✓

Solved JEE PYQ 2: Circular Motion (Centripetal Acceleration)

Question (JEE Main pattern): A particle moves in a circle of radius 0.5 m with a constant angular velocity of 4 rad/s. Find its linear speed and centripetal acceleration.

Step 1 — Linear speed:

v = ωr = 4 × 0.5 = 2 m/s

Step 2 — Centripetal acceleration:

a_c = ω²r = 4² × 0.5 = 16 × 0.5 = 8 m/s²

Cross-check with the other form: a_c = v²/r = 2²/0.5 = 4/0.5 = 8 m/s² ✓ (Unit check: (rad/s)² × m = m/s², since radian is dimensionless — consistent.)

Common PYQ Traps and Mistakes to Avoid

• Using v_x in vertical equations: Only u_y = u sinθ and v_y belong in vertical (g) equations. Mixing components is the number one error.
• Sign errors with g: If you take upward as positive, a_y = −g. Pick a sign convention at the start and stick to it.
• Confusing average velocity with average speed in a circle: Over a full revolution, average velocity = 0 (zero displacement), but average speed = 2πr/T ≠ 0.
• Forgetting θ = 45° only on level ground: The maximum-range angle changes for projection from a height or with wind.
• Thinking centripetal force is a new force: It is the net inward force supplied by tension, friction, gravity, or normal reaction.
• Using degrees in ω²r: Angular quantities must be in radians for v = ωr to hold.

Practice Questions with Answer Key

1. A projectile is launched at 45° with speed 20 m/s (g = 10 m/s²). Its maximum range R_max is:
   (a) 20 m (b) 40 m (c) 10 m (d) 80 m
2. Two projectiles are fired with the same speed at 20° and 70°. Their ranges are related as:
   (a) R₁ > R₂ (b) R₁ < R₂ (c) R₁ = R₂ (d) cannot say
3. A stone is thrown horizontally at 10 m/s from a 45 m high tower (g = 10 m/s²). It hits the ground after:
   (a) 3 s (b) 4.5 s (c) 1.5 s (d) 9 s
4. A particle moves in a circle of radius 2 m with speed 4 m/s. Its centripetal acceleration is:
   (a) 4 m/s² (b) 8 m/s² (c) 16 m/s² (d) 2 m/s²
5. In uniform circular motion, which quantity remains constant?
   (a) velocity (b) acceleration (c) speed (d) net force vector

Answer Key: 1 — (b) R_max = u²/g = 400/10 = 40 m; 2 — (c) 20° and 70° are complementary, so ranges are equal; 3 — (a) t = √(2h/g) = √(90/10) = 3 s; 4 — (b) a = v²/r = 16/2 = 8 m/s²; 5 — (c) speed.

Frequently Asked Questions

Q: Is horizontal velocity constant in projectile motion?

Yes. Ignoring air resistance, no horizontal force acts on the projectile, so v_x = u cosθ stays constant throughout the flight, while v_y changes by g each second.

Q: Why is centripetal acceleration needed even at constant speed?

Velocity is a vector. In circular motion the direction of the velocity changes continuously, so the velocity vector is changing — and a changing velocity means acceleration. This acceleration points toward the centre even when the speed is constant.

Q: For which two angles of projection is the range the same?

θ and 90° − θ (complementary angles) give equal range, since R = u² sin2θ/g and sin2θ = sin2(90° − θ). The range is maximum at θ = 45° for a given launch speed on level ground.

Q: How is projectile motion solved in JEE questions?

Resolve the initial velocity into x and y components (u cosθ and u sinθ), treat each axis as an independent 1D motion (uniform along x, uniformly accelerated along y), and use time t as the linking variable between the two equations.

Q: What is the relation between linear and angular velocity in circular motion?

v = ωr, where v is the linear (tangential) speed, ω is the angular velocity in rad/s, and r is the radius. Consequently, centripetal acceleration can be written as a = v²/r or equivalently as a = ω²r.

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