Physics Mechanics - How to Solve Projectile Motion Problems (Range, Max Height, Time of Flight) – IIT JEE Guide

How to Solve Projectile Motion Problems (Range, Max Height, Time of Flight) – IIT JEE Guide

Introduction

Mastering projectile motion unlocks 10+ marks in IIT JEE (Main + Advanced) and lets you predict everything from cricket sixes to rocket trajectories—without calculus. Miss this, and you lose easy marks on kinematics, vectors, and even circular motion problems.

WHAT YOU NEED TO KNOW FIRST

1. Vector resolution – Splitting velocity into horizontal (vₓ) and vertical (vᵧ) components.
2. Equations of motion – v = u + at, s = ut + ½at², v² = u² + 2as (for constant acceleration).
3. Acceleration due to gravity (g) – Acts downward (take g = 9.8 m/s² unless given otherwise; often rounded to 10 m/s² in JEE).

KEY TERMS & FORMULAS

Key Terms

Formulas (MEMORISE THESE – NOT GIVEN IN JEE SHEET!)

1. Time of Flight (T)

Formula: [ T = \frac{2 u \sin \theta}{g} ] Variables: - u = Initial velocity (m/s) - θ = Launch angle (degrees) - g = Acceleration due to gravity (9.8 m/s² or 10 m/s²)

Why? The projectile takes the same time to go up and come down.

2. Maximum Height (H)

Formula: [ H = \frac{u^2 \sin^2 \theta}{2g} ] Variables: Same as above.

Why? At max height, vertical velocity (vᵧ) becomes zero. Use v² = u² + 2as with v = 0, a = -g.

3. Range (R)

Formula: [ R = \frac{u^2 \sin 2\theta}{g} ] Variables: Same as above.

Why? Range depends on horizontal velocity (vₓ = u cos θ) and time of flight (T). Combine them: R = vₓ × T.

Special Case: For θ = 45°, sin 2θ = 1, so R = u²/g (maximum possible range).

4. Horizontal & Vertical Components of Velocity

Formulas: - vₓ = u cos θ (constant, no acceleration) - vᵧ = u sin θ - gt (changes due to gravity)

Why? Projectile motion is 2D motion—treat horizontal and vertical separately.

STEP-BY-STEP METHOD

Step 1: Draw a Diagram

• Sketch the projectile’s path (parabola).
• Label:
• Initial velocity (u) and angle (θ).
• Horizontal (x) and vertical (y) axes.
• Acceleration (g) acting downward.

Step 2: Resolve Initial Velocity into Components

• vₓ = u cos θ (horizontal, constant)
• vᵧ = u sin θ (vertical, changes)

Step 3: Find Time of Flight (T)

• Use: [ T = \frac{2 u \sin \theta}{g} ]
• Why? The time to go up = time to come down.

Step 4: Find Maximum Height (H)

• Use: [ H = \frac{u^2 \sin^2 \theta}{2g} ]
• Why? At max height, vᵧ = 0.

Step 5: Find Range (R)

• Use: [ R = \frac{u^2 \sin 2\theta}{g} ]
• Why? R = vₓ × T (horizontal velocity × time).

Step 6: Check for Special Cases

• If θ = 90° (straight up), T = 2u/g, H = u²/2g, R = 0.
• If θ = 45°, R = u²/g (maximum range).
• If θ = 30° or 60°, sin 2θ = √3/2 → R = (u²√3)/(2g).

Step 7: Verify Units & Plug in Values

• Ensure u is in m/s, θ in degrees, g in m/s².
• If g = 10 m/s², calculations simplify (common in JEE).

WORKED EXAMPLES

Example 1 – Basic (No Tricks)

Question: A ball is thrown with u = 20 m/s at θ = 30°. Find:
1. Time of flight (T)
2. Maximum height (H)
3. Range (R) (Take g = 10 m/s²)

Step 1: Draw Diagram

• u = 20 m/s, θ = 30°, g = 10 m/s².

Step 2: Resolve Velocity

• vₓ = u cos θ = 20 × cos 30° = 20 × (√3/2) = 10√3 m/s
• vᵧ = u sin θ = 20 × sin 30° = 20 × 0.5 = 10 m/s

Step 3: Time of Flight (T)

[ T = \frac{2 u \sin \theta}{g} = \frac{2 × 20 × 0.5}{10} = \frac{20}{10} = 2 \text{ s} ]

Step 4: Maximum Height (H)

[ H = \frac{u^2 \sin^2 \theta}{2g} = \frac{20^2 × (0.5)^2}{2 × 10} = \frac{400 × 0.25}{20} = \frac{100}{20} = 5 \text{ m} ]

Step 5: Range (R)

[ R = \frac{u^2 \sin 2\theta}{g} = \frac{20^2 × \sin 60°}{10} = \frac{400 × (√3/2)}{10} = \frac{200√3}{10} = 20√3 \text{ m} ]

What we did and why: - Split velocity into vₓ and vᵧ. - Used time of flight formula (since vᵧ becomes zero at max height). - Calculated range using vₓ × T. - No shortcuts—followed steps exactly.

Example 2 – Medium (Added Complication: Different Landing Height)

Question: A stone is thrown from a cliff 40 m high with u = 30 m/s at θ = 37°. Find:
1. Time of flight (T)
2. Range (R) (Take g = 10 m/s², sin 37° = 3/5, cos 37° = 4/5)

Step 1: Draw Diagram

• Initial height = 40 m (above ground).
• u = 30 m/s, θ = 37°, g = 10 m/s².

Step 2: Resolve Velocity

• vₓ = u cos θ = 30 × (4/5) = 24 m/s
• vᵧ = u sin θ = 30 × (3/5) = 18 m/s

Step 3: Time of Flight (T)

• The stone lands 40 m below the launch point.
• Use vertical motion equation: [ y = vᵧ t - \frac{1}{2} g t^2 ] [ -40 = 18t - 5t^2 ] [ 5t^2 - 18t - 40 = 0 ]
• Solve quadratic equation: [ t = \frac{18 ± \sqrt{(-18)^2 - 4 × 5 × (-40)}}{2 × 5} ] [ t = \frac{18 ± \sqrt{324 + 800}}{10} = \frac{18 ± \sqrt{1124}}{10} ] [ t = \frac{18 ± 33.5}{10} ]
• t = (18 + 33.5)/10 = 5.15 s (valid)
• t = (18 - 33.5)/10 = -1.55 s (invalid, discard)

Time of flight (T) = 5.15 s

Step 4: Range (R)

[ R = vₓ × T = 24 × 5.15 = 123.6 \text{ m} ]

What we did and why: - Landing height ≠ launch height → Modified vertical motion equation. - Solved quadratic equation for time (common in JEE). - Used vₓ × T for range (horizontal motion is always uniform).

Example 3 – Exam-Style (Disguised Problem)

Question (JEE 2018): A particle is projected from the ground with speed u at angle θ. If the maximum height is 3/8 of the range, find θ. (Take g = 10 m/s²)

Step 1: Write Given Formulas

• Max height: [ H = \frac{u^2 \sin^2 \theta}{2g} ]
• Range: [ R = \frac{u^2 \sin 2\theta}{g} ]

Step 2: Set Up the Equation

Given: [ H = \frac{3}{8} R ] Substitute: [ \frac{u^2 \sin^2 \theta}{2g} = \frac{3}{8} \left( \frac{u^2 \sin 2\theta}{g} \right) ]

Step 3: Simplify

• Cancel u²/g from both sides: [ \frac{\sin^2 \theta}{2} = \frac{3}{8} \sin 2\theta ]
• Recall: sin 2θ = 2 sin θ cos θ [ \frac{\sin^2 \theta}{2} = \frac{3}{8} × 2 \sin θ \cos θ ] [ \frac{\sin^2 \theta}{2} = \frac{3}{4} \sin θ \cos θ ]
• Divide both sides by sin θ (since sin θ ≠ 0): [ \frac{\sin θ}{2} = \frac{3}{4} \cos θ ]
• Rearrange: [ \tan θ = \frac{3/4}{1/2} = \frac{3}{2} ]
• Therefore: [ θ = \tan^{-1} \left( \frac{3}{2} \right) ]

What we did and why: - Disguised problem → Recognised H and R formulas. - Used trigonometric identities (sin 2θ = 2 sin θ cos θ). - Solved for θ using tan θ.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—projectile motion is free marks if you follow these steps:
1. Split velocity into vₓ = u cos θ (constant) and vᵧ = u sin θ (changes).
2. Time of flight is T = (2 u sin θ)/g—memorise it.
3. Max height is H = (u² sin² θ)/(2g)—vertical motion stops at the top.
4. Range is R = (u² sin 2θ)/g—double the angle, use sin 2θ.
5. If landing height ≠ launch height, solve the quadratic for time.
6. For 45°, range is maximum (R = u²/g).
7. Watch for g = 10 vs 9.8—don’t mix them up!
8. Draw a diagram—always. No excuses.

Now go crush those projectile problems!