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Kinematics & Projectile Motion: concepts, formulas & solved PYQ-style examples

Kinematics is the foundation of mechanics. Projectile questions are among the most frequently asked, and they become simple once you split motion into independent horizontal and vertical parts.

Horizontal: constant velocity. Vertical: constant acceleration g. Time links the two.

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Key concepts

Equations of motion

For constant acceleration only — v = u + at and the related forms.

Motion graphs

Slope of x–t is velocity, slope of v–t is acceleration, area under v–t is displacement.

Projectile motion

Time of flight, maximum height and range; complementary angles give equal range.

Horizontal projection

Thrown from a height: time depends only on the height.

Relative velocity

River–boat and rain–umbrella problems.

Formula sheet

Equations of motion

v=u+at,s=ut+12at2,v2=u2+2asv = u + at,\quad s = ut + \tfrac12 at^2,\quad v^2 = u^2 + 2as

Time of flight

T=2usin⁡θgT = \dfrac{2u\sin\theta}{g}

Maximum height

H=u2sin⁡2θ2gH = \dfrac{u^2\sin^2\theta}{2g}

Range

R=u2sin⁡2θgR = \dfrac{u^2\sin 2\theta}{g}

Horizontal projection from height h

t=2hg,x=u tt = \sqrt{\dfrac{2h}{g}},\quad x = u\,t

What PYQs keep asking

Patterns that repeat across previous-year JEE and NEET papers for this chapter.

  • Range, height and time of flight for a given speed and angle.
  • Relations such as “range equals four times maximum height”.
  • Ball thrown horizontally from a cliff or table — landing distance and impact velocity.
  • Reading displacement or acceleration from velocity–time graphs.

Solved examples

Example 1

A ball is projected at 20 m/s at 30° above the horizontal (g = 10 m/s²). Find the time of flight, maximum height and range.

  1. 1T=2usin⁡θg=2×20×0.510=2T = \dfrac{2u\sin\theta}{g} = \dfrac{2 \times 20 \times 0.5}{10} = 2 s.
  2. 2H=u2sin⁡2θ2g=400×0.2520=5H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{400 \times 0.25}{20} = 5 m.
  3. 3R=u2sin⁡60∘g=400×0.86610≈34.6R = \dfrac{u^2\sin 60^\circ}{g} = \dfrac{400 \times 0.866}{10} \approx 34.6 m.
Answer: T = 2 s, H = 5 m, R ≈ 34.6 m
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Example 2

A stone is thrown horizontally at 15 m/s from the top of a 45 m cliff (g = 10 m/s²). Find the time to land, the horizontal distance and the speed on impact.

  1. 1Vertical: t=2hg=9010=3t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{90}{10}} = 3 s.
  2. 2Horizontal: x=15×3=45x = 15 \times 3 = 45 m.
  3. 3At impact vx=15v_x = 15, vy=gt=30v_y = gt = 30 m/s, so v=152+302v = \sqrt{15^2 + 30^2}.
Answer: t = 3 s, x = 45 m, v ≈ 33.5 m/s
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Common traps

  • Using equations of motion when acceleration is not constant.
  • Using sin θ instead of sin 2θ in the range formula.
  • Thinking a horizontally thrown ball takes longer to fall than a dropped one — the fall time is the same.

Practise with

Projectile Motion CalculatorSUVAT / Kinematics CalculatorRotational MotionRay OpticsElectrostaticsProjectile motion simulation308 Kinematics practice questionsVideo solutions

FAQ

At what angle is the range of a projectile maximum?

At 45° on level ground (without air resistance); angles θ and 90° − θ give the same range.

Does horizontal speed affect fall time?

No. Fall time depends only on height and g; horizontal speed only changes how far the object lands.

Is kinematics important for JEE?

Directly it carries a few questions, but every mechanics chapter builds on it, so it matters far more than its own weightage.

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