Projectile on inclined plane — modified equations and trajectory analysis

Question

A projectile is launched from the base of an inclined plane (angle $\beta$ with horizontal) at an angle $\alpha$ with the incline. Derive the time of flight, range along the incline, and the condition for maximum range.

(JEE Advanced 2022 had a projectile-on-incline problem; JEE Main tests this 1-2 times per year)

Solution — Step by Step

Choose the right coordinate system

The trick: take axes along and perpendicular to the incline, not horizontal and vertical.

x-axis: along the incline (upward positive)
y-axis: perpendicular to the incline

In this frame, $g$ has two components:

Along incline: $g\sin\beta$ (decelerating the projectile along x)
Perpendicular to incline: $g\cos\beta$ (pulling it back to the surface)

Initial velocity components

If the launch speed is $u$ at angle $\alpha$ with the incline:

$u_x = u\cos\alpha, \quad u_y = u\sin\alpha$

The motion along the incline and perpendicular to it can now be treated independently.

Time of flight — when it returns to the incline

The projectile returns to the incline when $y = 0$ :

$0 = u\sin\alpha \cdot T - \frac{1}{2}g\cos\beta \cdot T^2$

$T = \frac{2u\sin\alpha}{g\cos\beta}$

Compare with flat ground: $T = 2u\sin\alpha/g$ . The denominator now has $g\cos\beta$ instead of $g$ .

Range along the incline

$R = u\cos\alpha \cdot T - \frac{1}{2}g\sin\beta \cdot T^2$

Substituting $T$ :

$R = \frac{2u^2\sin\alpha\cos(\alpha+\beta)}{g\cos^2\beta}$

Maximum range occurs when $\alpha = \frac{\pi}{4} - \frac{\beta}{2}$ , i.e., the launch angle bisects the angle between the incline and the vertical.

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Why This Works

By aligning the coordinate system with the incline, we reduce the problem to standard projectile motion — just with a modified gravitational acceleration. Along the incline, gravity has component $g\sin\beta$ (causing deceleration), and perpendicular to the incline, $g\cos\beta$ acts like "effective gravity."

The condition for maximum range ( $\alpha = 45° - \beta/2$ ) can be understood geometrically: the optimal launch angle bisects the angle between the incline surface and the vertical. On flat ground ( $\beta = 0$ ), this gives $\alpha = 45°$ , which is the familiar result.

Alternative Method

💡 Expert Tip

If you forget the incline formulas, you can always work in the standard horizontal-vertical frame. Set the incline equation as $y = x\tan\beta$ , write $x = u\cos(\alpha+\beta)\cdot t$ and $y = u\sin(\alpha+\beta)\cdot t - \frac{1}{2}gt^2$ , and find where the trajectory intersects the incline. This is longer but uses no new formulas.

Common Mistake

⚠️ Common Mistake

Students use the standard flat-ground formula $R = u^2\sin 2\alpha / g$ for inclined plane problems. This does not apply. The range formula changes because (1) gravity has a component along the incline, and (2) the "ground" is now tilted. Also, the maximum range angle is NOT 45° — it is $45° - \beta/2$ . Using 45° on an incline gives a wrong answer.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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