How to approach projectile motion problems — horizontal vs vertical decomposition

Question

A ball is thrown at $20$ m/s at an angle of $60°$ above the horizontal from ground level. Find (a) the time of flight, (b) the maximum height, and (c) the horizontal range. Take $g = 10$ m/s $^2$ .

(JEE Main & NEET standard problem)

Solution — Step by Step

Decompose the initial velocity

The entire trick of projectile motion is treating horizontal and vertical motion independently.

$u_x = u\cos\theta = 20\cos 60° = 10 \text{ m/s}$ $u_y = u\sin\theta = 20\sin 60° = 10\sqrt{3} \text{ m/s}$

Horizontal: no acceleration ( $a_x = 0$ ). Vertical: acceleration $= -g$ (downward).

Find the time of flight

The ball returns to ground level when vertical displacement = 0:

$0 = u_y t - \frac{1}{2}gt^2 = t\left(u_y - \frac{gt}{2}\right)$

So $t = 0$ (launch) or $T = \dfrac{2u_y}{g} = \dfrac{2 \times 10\sqrt{3}}{10} = \mathbf{2\sqrt{3} \approx 3.46 \text{ s}}$

Find the maximum height

At the highest point, $v_y = 0$ . Using $v_y^2 = u_y^2 - 2gH$ :

$H = \frac{u_y^2}{2g} = \frac{(10\sqrt{3})^2}{2 \times 10} = \frac{300}{20} = \mathbf{15 \text{ m}}$

Find the horizontal range

$R = u_x \times T = 10 \times 2\sqrt{3} = \mathbf{20\sqrt{3} \approx 34.6 \text{ m}}$

Or use the range formula directly: $R = \dfrac{u^2\sin 2\theta}{g} = \dfrac{400 \times \sin 120°}{10} = \dfrac{400 \times \sqrt{3}/2}{10} = 20\sqrt{3}$ m.

Why This Works

Gravity only acts vertically. This means horizontal velocity never changes — the ball moves at constant speed horizontally. Vertically, it's just a ball thrown straight up and coming back down.

By decomposing into two independent 1D problems, we turn a curved path into two straight-line motions. Time is the link between them — whatever time the vertical motion gives us, the horizontal motion uses the same time.

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Alternative Method — Using the Trajectory Equation

For problems where you need height at a specific horizontal distance $x$ :

$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$

This eliminates time entirely. Useful when the question gives $x$ and asks for $y$ , or vice versa.

💡 Expert Tip

Complementary angles ( $\theta$ and $90° - \theta$ ) give the same range but different heights and flight times. JEE loves asking: "Two projectiles give the same range. If one is at $30°$ , what's the other angle?" Answer: $60°$ . The steeper angle flies higher and longer.

Common Mistake

⚠️ Common Mistake

Students apply the range formula $R = u^2\sin 2\theta / g$ when the projectile doesn't land at the same height it was launched from (e.g., thrown from a cliff). This formula only works for ground-to-ground projectiles. For unequal heights, you must solve the quadratic $y = u_y t - \frac{1}{2}gt^2$ with $y \neq 0$ . JEE Main 2024 had a cliff-launch problem — many students used the wrong formula.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Trajectory Equation

Common Mistake

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