How to approach projectile motion problems — horizontal vs vertical decomposition

Question

How do we systematically solve projectile motion problems by decomposing motion into horizontal and vertical components? What formulas apply to each direction?

(CBSE 11, JEE Main, NEET — projectile motion is among the most frequently tested kinematics topics)

Solution — Step by Step

The golden rule — independence of motions

Projectile motion is just two independent motions happening simultaneously:

Horizontal ( $x$ -direction): Uniform velocity (no acceleration, ignoring air resistance)
Vertical ( $y$ -direction): Uniformly accelerated motion under gravity ( $g = 9.8$ m/s $^2$ downward)

These two motions share only one variable: time ( $t$ ). This is the link between them.

Initial velocity decomposition

If a projectile is launched with velocity $u$ at angle $\theta$ to the horizontal:

Horizontal component: $u_x = u\cos\theta$ (remains constant throughout)
Vertical component: $u_y = u\sin\theta$ (changes due to gravity)

Horizontal equations: $x = u\cos\theta \cdot t$

Vertical equations: $v_y = u\sin\theta - gt$ $y = u\sin\theta \cdot t - \frac{1}{2}gt^2$ $v_y^2 = u^2\sin^2\theta - 2gy$

Key results for standard projectile

Projectile Motion Formulas

Time of flight: $T = \frac{2u\sin\theta}{g}$

Maximum height: $H = \frac{u^2\sin^2\theta}{2g}$

Range: $R = \frac{u^2\sin 2\theta}{g}$

Maximum range occurs at $\theta = 45°$ : $R_{max} = \frac{u^2}{g}$

Complementary angle property: $\theta$ and $(90° - \theta)$ give the same range but different heights and times of flight. For example, $30°$ and $60°$ give equal range.

Problem-solving algorithm

Set up coordinates: $x$ horizontal, $y$ vertical upward
Decompose initial velocity into $u_x$ and $u_y$
Identify what is asked (time, height, range, velocity at a point)
Use vertical equations to find time (usually from a height condition)
Use that time in horizontal equations to find range or position
If asked for velocity: find $v_x$ and $v_y$ separately, then $v = \sqrt{v_x^2 + v_y^2}$

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Why This Works

The decomposition works because gravity acts only vertically — it has no horizontal component. So the horizontal motion is completely unaffected by gravity (constant velocity), while the vertical motion is just free fall with an initial vertical velocity. By treating them separately and connecting through time, we convert a 2D problem into two simpler 1D problems.

This is a direct consequence of Newton's second law applied component-wise: $F_x = 0$ (so $a_x = 0$ ) and $F_y = -mg$ (so $a_y = -g$ ).

Common Mistake

⚠️ Common Mistake

The most frequent error: using $u$ (total velocity) instead of $u\sin\theta$ (vertical component) in the height formula. The formula $H = u^2\sin^2\theta / 2g$ uses the vertical component. Students who write $H = u^2/2g$ forget the $\sin^2\theta$ factor and overestimate the height. Similarly, time of flight depends on $u\sin\theta$ , not $u$ . Always decompose first, calculate second.

💡 Expert Tip

At maximum height, the vertical velocity is zero but the horizontal velocity is still $u\cos\theta$ . So the speed at the highest point is $u\cos\theta$ , NOT zero. JEE Main tests this almost every year — "What is the speed at the highest point?" Answer: $u\cos\theta$ .

Question

Solution — Step by Step

Why This Works

Common Mistake

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