PhysicsMotion in a planeClass 11

Projectile motion: interactive simulation

A projectile moves under gravity alone: its horizontal velocity never changes, while its vertical velocity drops by g every second. Put those two independent motions together and you get a parabola.

Useful for: JEE Main, NEET, AP Physics 1

Try this

Drag the launch arrow, change the gravity and raise the launch point. The range, height and time of flight are calculated from the exact equations as you move — the same ones you use in the exam.

  1. 1

    Keep u = 20 m/s and g = 10. Compare θ = 30° with θ = 60°.

    What you should see

    Both land at R ≈ 34.6 m, but 60° climbs to 15 m (vs 5 m) and stays in the air 3.46 s (vs 2 s).

  2. 2

    Find the angle that gives the longest range on level ground.

    What you should see

    45°, where R = u²/g = 40 m for u = 20 m/s and g = 10.

  3. 3

    Raise the launch height to 20 m and search for the best angle again.

    What you should see

    The best angle is now less than 45° — the extra fall time rewards a flatter throw.

  4. 4

    Press Launch and watch the green and red arrows on the ball.

    What you should see

    Green (horizontal velocity) never changes. Red (vertical velocity) shrinks to zero at the top, then grows downward.

  5. 5

    Switch gravity from Earth to the Moon without changing u or θ.

    What you should see

    T, H and R all grow by 9.8 / 1.62 ≈ 6 times, because each is proportional to 1/g.

Formulas behind it

Time of flight (level ground)

T=2usin⁡θgT = \frac{2u\sin\theta}{g}

Maximum height

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

Horizontal range

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

Equation of the path

y=xtan⁡θ−g x22u2cos⁡2θy = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\theta}

Launched from height h

T=usin⁡θ+u2sin⁡2θ+2ghgT = \frac{u\sin\theta + \sqrt{u^2\sin^2\theta + 2gh}}{g}

Common mistakes

  • Writing sin θ instead of sin 2θ in the range formula.
  • Saying the speed is zero at the top. Only the vertical component is zero; the speed there is u cos θ.
  • Using 45° for maximum range when the projectile is launched from a height — that rule is only for level ground.
  • Using T = 2u sin θ / g when the launch and landing heights are different.
Practise 308 questionsCheck your answer instantly, with a worked solution.Kinematics & Projectile Motion chapter guideFormula sheet, solved examples and PYQ patterns.Stuck on a projectile motion question?Snap it and get a whiteboard video solution.

FAQ

Why do 30° and 60° give the same range?

Range depends on sin 2θ, and sin 60° = sin 120°. Any two angles that add up to 90° give the same range on level ground, but the steeper one goes higher and takes longer.

At what angle is the maximum height equal to the range?

H/R = tan θ / 4, so H = R when tan θ = 4, i.e. θ ≈ 76°.

Does the mass of the ball change the path?

No. Without air resistance every object has the same acceleration g, so mass cancels out of every formula.

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