PhysicsPHY 02
Kinematics & Projectile Motion
Equations of motion, projectile, relative velocity, displacement-velocity-acceleration graphs
Concept Core
Essential theory — everything NCERT tests on Kinematics & Projectile Motion
EQUATIONS OF MOTION
For uniform acceleration: v = u + at; s = ut + ½at²; v² = u² + 2as; sn = u + a(2n−1)/2
Free fall: a = g = 9.8 m/s² (downward). At maximum height: v = 0. Time of flight for vertical throw: t = 2u/g.
PROJECTILE MOTION
Horizontal: uₓ = u cosθ, aₓ = 0, x = uₓt
Vertical: uᵧ = u sinθ, aᵧ = −g, y = uᵧt − ½gt²
Time of flight: T = 2u sinθ/g. Maximum height: H = u²sin²θ/(2g). Range: R = u²sin2θ/g. Maximum range at θ = 45°.
RELATIVE VELOCITY
vAB = vA − vB (velocity of A relative to B)
Rain-man problem: angle depends on man's velocity. vrain w.r.t. man = vrain − vman.
Two cars approach each other: relative speed = v₁ + v₂. Same direction: |v₁ − v₂|.
GRAPHS
s-t graph: slope = velocity. Horizontal line = rest. Parabola = uniform acceleration.
v-t graph: slope = acceleration. Area = displacement. Horizontal line = uniform velocity. Triangle area = displacement in uniform acceleration.
a-t graph: area = change in velocity.
IMPORTANT RESULTS
Complementary angles (θ and 90°−θ) give same horizontal range. Max range = u²/g (at 45°).
Horizontal projectile: time to hit ground = √(2h/g). Range = u√(2h/g).
Height H = R tan θ/4 relation for range and height.
Fact & Formula Vault
High-yield facts, numbers, and formulas
Kinematic Equations
v = u + at
s = ut + ½at²
v² = u² + 2as
sₙ = u + a(2n−1)/2
Projectile Formulae
T = 2u sinθ/g
H = u²sin²θ/2g
R = u²sin2θ/g
R_max = u²/g at 45°
Graph Slopes
s-t slope = velocity
v-t slope = acceleration
v-t area = displacement
a-t area = Δvelocity
Worked Examples
NEET-style questions solved step-by-step
EASYA ball is thrown at 30° with speed 20 m/s. Maximum height?▾
A ball is thrown at 30° with speed 20 m/s. Maximum height?
H = u²sin²θ/(2g) = 20² × sin²30°/(2×10) = 400 × 0.25/20 = 5 m
MEDIUMTwo trains approach each other at 60 km/h and 80 km/h. Relative speed?▾
Two trains approach each other at 60 km/h and 80 km/h. Relative speed?
Moving toward each other → relative speed = 60 + 80 = 140 km/h. If same direction: |80−60| = 20 km/h.
HARDFor what two angles is the range of a projectile equal if speed is same?▾
For what two angles is the range of a projectile equal if speed is same?
Complementary angles: θ and (90°−θ). E.g., 30° and 60° give same range since sin2(30°)=sin2(60°)=sin120°=sin60°=√3/2.
Mistake DNA
Common NEET traps for this chapter
⚠ Range formula angle
R = u²sin2θ/g. Maximum at 2θ = 90° → θ = 45°. NOT at 90° (that gives zero range — goes straight up).
✓ Fix: Max range = u²/g at 45°
⚠ sₙ formula
sₙ = u + a(2n−1)/2 gives displacement in n-th second only, not total.
✓ Fix: sₙ = distance in n-th second; s = total after n seconds
⚠ v-t graph area
Area under v-t graph = displacement (can be negative). NOT distance (distance = area counting all as positive).
✓ Fix: Area = displacement. |Area| = distance if direction doesn't change
Chapter Intelligence
Exam data and last-minute strategy
NEET Frequency
3–4 Q/year. Projectile formulae (range, height, time), graph interpretation, equations of motion are standard every year.
High-Yield
R = u²sin2θ/g. R_max at 45°. Complementary angles: same range. T = 2u sinθ/g. H = u²sin²θ/2g. v-t area = displacement.
Strategy
Memorise all 4 projectile formulae. Practice graph reading. These are formula-application questions — speed through elimination.
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