Projectile on an Inclined Plane (Class 11)

In normal projectile motion, the object lands on horizontal ground.

But here:

• The projectile is thrown

• It lands on a sloping surface (inclined plane)

Let the plane make an angle β with the horizontal.

📌 Setup of the Problem

• Inclined plane angle = β

• Projectile is thrown with speed = u

• Angle of projection (from horizontal) = θ

• Acceleration due to gravity = g

⚙️ Step 1: Resolve Initial Velocity

Horizontal component:

ux=ucos⁡θu_x = u \cos\thetaux=ucosθ

Vertical component:

uy=usin⁡θu_y = u \sin\thetauy=usinθ

⚙️ Step 2: Motion Equations

Horizontal motion:

x=ucos⁡θ⋅tx = u\cos\theta \cdot tx=ucosθ⋅t

Vertical motion:

y=usin⁡θ⋅t−12gt2y = u\sin\theta \cdot t - \frac{1}{2}gt^2y=usinθ⋅t−21gt2

📌 Condition to Hit the Inclined Plane

Equation of inclined plane:

y=xtan⁡βy = x \tan\betay=xtanβ

At the point of landing:

usin⁡θ⋅t−12gt2=(ucos⁡θ⋅t)tan⁡βu\sin\theta \cdot t - \frac{1}{2}gt^2 = (u\cos\theta \cdot t)\tan\betausinθ⋅t−21gt2=(ucosθ⋅t)tanβ

Solve for time ttt.

🕒 Time of Flight on Inclined Plane

T=2usin⁡(θ−β)gcos⁡βT = \frac{2u \sin(\theta - \beta)}{g\cos\beta}T=gcosβ2usin(θ−β)

(Important formula for exams)

📏 Range Along the Inclined Plane

Range measured along the plane is:

R=2u2cos⁡θsin⁡(θ−β)gcos⁡2βR = \frac{2u^2 \cos\theta \sin(\theta - \beta)}{g\cos^2\beta}R=gcos2β2u2cosθsin(θ−β)

🎯 Condition for Maximum Range

Maximum range on inclined plane occurs when:

θ=45∘+β2\theta = 45^\circ + \frac{\beta}{2}θ=45∘+2β

Very important result for Class 11 exams.

🧠 Key Idea to Remember

In inclined plane problems:

• Resolve motion normally (horizontal & vertical)

• Apply condition of plane y=xtan⁡βy = x\tan\betay=xtanβ

• Solve carefully

This concept is an extension of the projectile motion studied by
Galileo Galilei.

🔥 Quick Summary