Projectile motion is the most-tested kinematics concept in JEE Main.
Every year, 1-2 questions come directly from projectile motion. The key is understanding that horizontal and vertical motion are independent — gravity only affects the vertical component.
Projectile Motion Formulas
Range
R = u²sin2θ/g
Maximum at θ = 45°
Max Height
H = u²sin²θ/2g
Maximum at θ = 90°
Time of Flight
T = 2u·sinθ/g
Independent of horizontal motion
Horizontal Velocity
vx = u·cosθ (constant)
Gravity doesn't affect vx
How to solve projectile motion problems in JEE
- Resolve initial velocity — v₀ₓ = u cos θ, v₀ᵧ = u sin θ. Horizontal and vertical components are independent.
- Choose axis/sign convention — upward = +, rightward = +. Gravity g = -9.8 m/s² in vertical direction.
- Apply kinematic equations — x = v₀ₓt, y = v₀ᵧt + ½gt², vᵧ = v₀ᵧ + gt. Use time t consistently.
- For max range — set θ = 45° — R = u²sin2θ/g is max when sin2θ = 1, so 2θ = 90°, θ = 45°.
- Check if launched from height — use y = v₀ᵧt - ½gt² and solve for t when y = -h.
What students get wrong
❌ Forgetting vx is constant
Horizontal velocity never changes (no air resistance). Only vy changes due to gravity.
❌ Using 45° for max height
45° gives max RANGE. 90° gives max HEIGHT. These are different!
❌ Mixing up sin and cos
Range uses sin2θ. Height uses sin²θ. Time uses sinθ. One letter difference changes everything.
❌ Ignoring the sign of g
g is negative when the projectile moves up, positive when it comes down. Set up your sign convention first.
Projectile Motion Questions
Why is 45° the angle for maximum range?
Range R = u²sin2θ/g. sin2θ is maximum when 2θ = 90°, so θ = 45°. At this angle, the horizontal and vertical components of velocity are perfectly balanced.
Does mass affect projectile motion?
No! Without air resistance, mass doesn't appear in any projectile formula. A feather and a cannonball follow the same path in a vacuum.
What if the projectile is launched from a height?
Then you can't use the simple range formula. You need to solve the vertical equation s = ut + ½at² with the initial height, then find when y = 0.
Is horizontal velocity really constant?
Yes, in ideal projectile motion (no air resistance). Gravity acts only vertically, so vx = u·cosθ stays the same throughout the flight.