Projectile Motion — The Battlefield
Projectile Motion
The Battlefield
A warrior who knows the trajectory controls the battlefield. In JEE, the trajectory is the question.
The Launch
Choosing your angle of attack
The Launch
Choosing your angle of attack
The Peak
The high ground — maximum advantage
The Peak
The high ground — maximum advantage
The Range
How far your arrow reaches
The Range
How far your arrow reaches
The Landing
Impact — symmetry of war
The Landing
Impact — symmetry of war
Core Equations
Time of Flight
T = 2u sin θ / g
Total airtime
Maximum Height
H = u² sin²θ / 2g
Highest point
Range
R = u² sin 2θ / g
Horizontal distance
Max Range
R_max = u²/g at θ = 45°
When sin 2θ = 1
Component Equations
Horizontal
x = (u cos θ)t
No acceleration
Vertical position
y = (u sin θ)t - ½gt²
Parabolic path
Vertical velocity
vᵧ = u sin θ - gt
Decreases, then reverses
Trajectory equation
y = x tan θ - gx²/(2u²cos²θ)
Eliminate t
Standard Projectile Common
Find T, H, R given u and θ
Direct formula substitution. Convert degrees to check sin/cos values.
Using sin θ where sin 2θ is needed for range. Double angle!
Complementary angle problems
sin 2(90°-θ) = sin(180°-2θ) = sin 2θ. Same range.
Assuming same range means same height. Heights are DIFFERENT for complementary angles.
Projectile from Height Common
Thrown horizontally from cliff
θ = 0, so u_y = 0. Time from h = ½gt². Range = ut.
Adding initial height to max height formula. If thrown horizontally, there IS no max height above launch.
Thrown at angle from building top
Set y = -h (below launch). Solve quadratic in t. Take positive root.
Sign convention — if down is negative, initial height gives y₀ = +h and ground is y = 0.
JEE Advanced Patterns Moderate
Two projectiles — collision or meeting point
Set x₁ = x₂ and y₁ = y₂. Solve for t. Check if valid (t > 0 for both).
Forgetting that both projectiles must be in flight at the meeting time.
Projectile on inclined plane
Take axes along and perpendicular to incline. g splits into g sin α (along) and g cos α (perpendicular).
Using standard formulas without modifying for the incline. The 'ground' is tilted — redefine your axes.
Find angle for given range (R < R_max)
R = u² sin 2θ / g → sin 2θ = Rg/u². Two solutions: θ and (90°-θ).
Giving only one angle. There are always TWO angles for R < R_max.
| Property | At θ = 30° | At θ = 45° | At θ = 60° |
|---|---|---|---|
| Range R | u²√3/(2g) | u²/g (MAX) | u²√3/(2g) |
| Max Height H | u²/(8g) | u²/(4g) | 3u²/(8g) |
| Time of Flight T | u/g | u√2/g | u√3/g |
| Same range as | 60° | Only 45° | 30° |
JEE Traps
1. Using R = u²sin θ/g instead of sin 2θ
2. Saying velocity at top = 0 (only vertical component is zero!)
3. Forgetting that horizontal velocity NEVER changes
4. Not considering two angles for same range
Memory Fixes
1. Range has '2θ' — always double the angle
2. At top: v = u cos θ (not zero!). Only vᵧ = 0
3. Horizontal: no force → no acceleration → constant velocity
4. Two angles: θ and (90°-θ). Always give both.
'What is the velocity at the highest point?'
Answer is NOT zero. It is u cos θ (horizontal component). Vertical component is zero, but the projectile is still moving horizontally. This question appears every 2-3 years.
Range formula?
6 cards — tap to flip
A ball is thrown at 60° and another at 30° with the same speed u. Which statement is TRUE?
Complementary angles (30° + 60° = 90°) give the same range (R = u²sin2θ/g, and sin 60° = sin 120°). But heights differ: H₆₀ = 3u²/8g vs H₃₀ = u²/8g.