CLASS 11 · CHAPTER 3 · MECHANICS

Motion in a Plane

Any object launched into the air and left to gravity alone, a ball, a shell, a shot put, follows the same curve. Here's why.

01 · See it

Watch it happen

Want the full chapter, section by section, with vectors, resolution, and uniform circular motion all spelled out? Read the detailed notes →

Launch speed: 28 m/sLaunch angle: 45°

Compare with the complementary angle (45°): same range, notice?

80.0 m

Range

20.0 m

Max height

4.04 s

Time of flight

The horizontal and vertical motions are independent. Horizontally, nothing pushes or slows the object; it just carries its launch speed at a constant rate. Vertically, gravity alone decides everything. Drag the sliders and watch both truths hold at every angle.

02 · Derive it

Where the formula comes from

Split the launch velocity $v_0$ at angle $\theta$ into components:

\begin{gathered} v_x = v_0\cos\theta \quad(\text{constant}) \\[6px] v_y = v_0\sin\theta - gt \end{gathered}

Integrating position from these:

\begin{gathered} x(t) = v_0\cos\theta \cdot t \\[6px] y(t) = v_0\sin\theta \cdot t - \tfrac{1}{2}gt^2 \end{gathered}

Time of flight comes from solving $y(t) = 0$ for the second time (the landing):

Time of flight

\textcolor{#e08a1e}{T} = \dfrac{2v_0\sin\theta}{g}

Max height is where $v_y = 0$ , substituted back into $y(t)$ . Range is $x(T)$ :

Max height and range

\begin{gathered} \textcolor{#e08a1e}{H} = \dfrac{v_0^2\sin^2\theta}{2g} \\[8px] \textcolor{#e08a1e}{R} = \dfrac{v_0^2\sin 2\theta}{g} \end{gathered}

03 · Break it

Where the shortcut stops working

Every CBSE and JEE student learns “time to go up equals time to come down.” Look at the derivation again: that split only falls out of the formula for $T$ because we assumed the object lands at the same height it launched from, the two roots of a symmetric parabola.

Launch a ball off a cliff, or from a table onto the floor, and the symmetry is gone. The ball spends longer falling than rising, because it has further to fall than it rose. The formula $T = 2v_0\sin\theta/g$ simply does not apply; you have to solve the full quadratic $y_0 + v_0\sin\theta\cdot t - \tfrac{1}{2}gt^2 = 0$ for the actual landing height.

Same story with range at 45°: it’s only the maximum when launch and landing heights match. Launch from a height, and the optimal angle is always less than 45°, because the projectile gets more benefit from covering ground quickly than from staying airborne longer.

04 · Master it

Apply it under exam conditions

Q1. A ball is thrown at 20 m/s at 30° above the horizontal. Find its range and maximum height. (g = 9.8 m/s²)

R = \dfrac{v_0^2\sin2\theta}{g} = \dfrac{400\times\sin60^\circ}{9.8} \approx \textcolor{#e08a1e}{35.3\text{ m}}

H = \dfrac{v_0^2\sin^2\theta}{2g} = \dfrac{400\times\sin^230^\circ}{19.6} \approx \textcolor{#e08a1e}{5.1\text{ m}}

Q2. Two angles give the same range for a given speed. If one is 20°, what’s the other, and why?

70°. Since $\sin2\theta = \sin(180^\circ - 2\theta)$ , complementary angles (θ and 90°−θ) always share a range: one is a “flat and fast” trajectory, the other “steep and slow,” covering the same ground.

05 · FAQs

Quick answers

Why is horizontal velocity constant during projectile motion?+

Because gravity acts only vertically, there is no horizontal force at all once air resistance is ignored, so by Newton's first law the horizontal velocity simply never changes.

Does the mass of the projectile affect its range or time of flight?+

No. Every term in the range and time-of-flight formulas comes from kinematics, not force, so mass cancels out entirely. A heavier and a lighter object launched the same way follow identical paths, ignoring air resistance.

What's the actual shape of a projectile's path?+

A parabola. y(t) is quadratic in t while x(t) is linear in t, and eliminating t between the two always produces a quadratic relationship between y and x.

Is 45° really always the best launch angle for maximum range?+

Only when launch and landing heights are equal. Launch from any height above the landing point and the optimal angle is always less than 45°.

Why do two different launch angles give the same range?+

Range depends on sin(2θ), and sine of complementary angles (2θ and 180°−2θ) are equal, so θ and 90°−θ always produce the same range for a given launch speed.

Related concepts

Physics doesn’t stay inside chapter boundaries. Neither should you.

Mechanics

Motion in a Straight Line

Scrub through time and watch position, velocity, and the area under the v-t graph move together.

Mechanics

Laws of Motion

Push a block and watch friction switch from static to kinetic the instant it slips.