Projectile Motion: Trajectory, Time of Flight, Range

A projectile launched at an angle follows a parabolic path. Learn the trajectory equation, time of flight, maximum height and range — the four results that power a quarter of JEE kinematics questions.

Projectile Motion1 min · Free lecture

In this lesson

Derive the trajectory equation of a projectile
Compute time of flight, maximum height and range from launch speed and angle
Apply the results to numerical JEE-style problems

A projectile is any body launched with an initial velocity and left to move under gravity alone. We ignore air resistance and take $g$ constant. The launch velocity $v_0$ at angle $\theta$ to the horizontal splits into $v_{0x} = v_0\cos\theta$ and $v_{0y} = v_0\sin\theta$ . Gravity acts only on the vertical component, so the horizontal motion is uniform and the vertical motion is uniformly accelerated — that single split is the whole trick.

x = v_0 \cos\theta \, t, \qquad y = v_0 \sin\theta \, t - \tfrac{1}{2} g t^2

Position of a projectile at time t (origin at launch point)

v_0 — launch speed (m/s)
\theta — launch angle above horizontal (rad)
g — acceleration due to gravity (m/s^2)
t — time since launch (s)

y = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta}

Trajectory equation — y as a function of x (a parabola)

x — horizontal distance from launch (m)
y — height above launch point (m)

The trajectory equation is a quadratic in $x$ with a negative coefficient — hence the parabolic path. Three derived results follow directly from the kinematics and are memorised by every JEE aspirant.

T = \frac{2 v_0 \sin\theta}{g}, \qquad H = \frac{v_0^2 \sin^2\theta}{2g}, \qquad R = \frac{v_0^2 \sin 2\theta}{g}

Time of flight T, maximum height H, and horizontal range R

T — total time of flight (s)
H — maximum height reached (m)
R — horizontal range (back to launch level) (m)

Range is maximum at 45°, not at bigger angles. And

\sin 2\theta

is symmetric: angles

\theta

and

90° - \theta

give the same range — a classic JEE trap.

Worked example

A projectile is launched at 40 m/s at 30° above horizontal. Take g = 10 m/s². Find (a) time of flight, (b) maximum height, (c) range.

T = 2 v0 sinθ / g = 2 × 40 × sin 30° / 10 = 2 × 40 × 0.5 / 10 = 4 s
H = v0² sin²θ / 2g = 1600 × 0.25 / 20 = 20 m
R = v0² sin 2θ / g = 1600 × sin 60° / 10 = 1600 × 0.866 / 10 ≈ 138.6 m

Answer: T = 4 s, H = 20 m, R ≈ 138.6 m

Animated figure: projectile-trajectory (rendered in video)

The parabolic path with v0, θ, H, R and the velocity split at launch

Transcript (1 min)

WEBVTT

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Projectile Motion — Trajectory, Time of Flight, Range
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v0 splits into v0 cos θ (horizontal) and v0 sin θ (vertical)
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Position: x = v0 cos θ t, y = v0 sin θ t − ½ g t²
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Trajectory: y = x tan θ − g x² / (2 v0² cos² θ)
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T = 2 v0 sin θ / g · H = v0² sin² θ / 2g · R = v0² sin 2θ / g
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Example: v0 = 40 m/s, θ = 30°, g = 10 → T = 4 s, H = 20 m, R ≈ 138.6 m
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00:01:14.000 --> 00:01:22.000
Recap: split velocity → uniform horizontal + accelerated vertical

Frequently asked

Why is the range of a projectile maximum at 45 degrees?

The range is R = v0² sin(2θ)/g. For a fixed launch speed, R is largest when sin(2θ) is maximum, which happens at 2θ = 90°, i.e. θ = 45°. Angles θ and 90° − θ give equal ranges.

Does the time of flight depend on the horizontal component of velocity?

No. Time of flight is T = 2 v0 sin(θ)/g — it depends only on the vertical launch component and gravity. The horizontal component affects only how far the projectile travels, not how long it stays in the air.

When is the trajectory equation y = x tan θ − gx²/(2v0²cos²θ) valid?

When the launch and landing points are at the same height, air resistance is neglected, and g is constant. For projectiles launched from a height, the landing point shifts and the range formula changes.

Printable notes (free)

Practice

3 free · 1 premium

Q1 · Numerical · difficulty 2/5

A projectile is launched with speed 15 m/s at an angle 70° above the horizontal. Taking g = 10 m/s², find its horizontal range.

R = \frac{v_0^2 \sin 2\theta}{g}

Hint available · solution with Premium.

Q2 · Numerical · difficulty 2/5

A projectile is launched with speed 30 m/s at an angle 30°. Taking g = 9.8 m/s², find its maximum height above the launch point.

H = \frac{v_0^2 \sin^2\theta}{2g}

Hint available · solution with Premium.

Q3 · MCQ · difficulty 1/5

For a fixed launch speed, the horizontal range of a projectile is maximum when the launch angle is:

30°
45°
60°
90°

Hint available · solution with Premium.

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