A plane is inclined at an angle alpha = 30 degrees with respect to the horizontal. A particle is pro

Q: Why do we tilt the coordinate axes?

Tilting the axes aligns the X-axis with the surface of the incline. This means the final Y-displacement when the particle hits the plane is exactly zero, which makes calculating the time of flight incredibly simple.

Q: Why is the acceleration in the X-direction negative?

We defined the positive X-direction as 'up the ramp'. Gravity pulls straight down, so its component along the ramp points 'down the ramp', opposing the initial motion. Hence, $a_x = -g \sin\alpha$.

Q: How do we calculate $\sin 15^\circ$ without a calculator?

You can use the angle subtraction formula: $\sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$. This gives $\frac{\sqrt{3} - 1}{2\sqrt{2}}$, which is approximately $0.2588$.

Q: Why does the time of flight only depend on the Y-components?

Because the 'ground' in our tilted system is the X-axis. The particle leaves the X-axis and returns to it. The time it takes to do this is governed entirely by its initial velocity away from the axis ($u_y$) and the acceleration pulling it back ($a_y$).

Q: What happens if the angle of projection $\theta$ is measured from the horizontal instead of the plane?

If $\theta$ was measured from the horizontal, the angle of projection relative to the inclined plane would be $(\theta - \alpha)$. You would just substitute $(\theta - \alpha)$ wherever you see $\theta$ in the tilted-axis formulas.

Question

A plane is inclined at an angle $\alpha = 30^{\circ}$ with respect to the horizontal. A particle is projected with a speed $u = 2 	ext{ ms}^{-1}$, from the base of the plane, making an angle $	heta = 15^{\circ}$ with respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane is close to
[Take, $g = 10 	ext{ ms}^{-2}$]

Options:

(A) 26 cm

(B) 20 cm

(C) 18 cm

(D) 14 cm

Sigma Percentile · Accepted Answer

### The Tilted World of Inclined Planes

Imagine standing on a steep ramp and throwing a ball upwards. The standard horizontal and vertical coordinate system suddenly feels clunky. Gravity is pulling straight down, but the ball is moving along a slanted surface.

The most elegant trick in physics for this scenario is to **tilt your perspective**. By rotating our coordinate axes so that the $X$-axis runs parallel to the inclined plane and the $Y$-axis is perpendicular to it, we transform a complex 2D problem into two simple, independent 1D motions.

### Breaking Down the Vectors

Once we tilt our axes, we must resolve both our initial velocity $u$ and the acceleration due to gravity $g$ into these new $X$ and $Y$ components.

The particle is launched at $u = 2 	ext{ ms}^{-1}$ at an angle $	heta = 15^\circ$ relative to the plane. 
- The velocity along the plane is $u_x = u \cos	heta = 2 \cos 15^\circ$.
- The velocity perpendicular to the plane is $u_y = u \sin	heta = 2 \sin 15^\circ$.

Now, look at gravity. It points straight down, but in our tilted frame, it drags the particle back down the ramp and pulls it into the surface. 
- The acceleration along the plane is $a_x = -g \sin\alpha = -10 \sin 30^\circ$.
- The acceleration perpendicular to the plane is $a_y = -g \cos\alpha = -10 \cos 30^\circ$.

Notice the **negative signs**! Both components of gravity are acting against the positive directions of our tilted axes.

### The Time of Flight

How long does the particle stay in the air? This depends entirely on the $Y$-motion. The particle leaves the plane ($Y=0$), travels upwards, and eventually hits the plane again ($Y=0$).

Using the kinematic equation $S_y = u_y T + \frac{1}{2} a_y T^2$ and setting $S_y = 0$, we get the time of flight formula:

$$T = \frac{2u_y}{|a_y|} = \frac{2u \sin	heta}{g \cos\alpha}$$

Plugging in our values:

$$T = \frac{2(2) \sin 15^\circ}{10 \cos 30^\circ} = \frac{4 	imes 0.2588}{10 	imes 0.866} \approx 0.119 	ext{ s}$$

### The Final Range

Now that we know the particle is airborne for $0.119$ seconds, we can find out how far it traveled along the $X$-axis (the ramp). We use the second equation of motion for the $X$-direction:

$$R = u_x T + \frac{1}{2} a_x T^2$$

Substitute the values carefully, remembering that $a_x$ is negative because gravity is slowing the particle's ascent up the ramp:

$$R = (2 \cos 15^\circ)(0.119) - \frac{1}{2}(10 \sin 30^\circ)(0.119)^2$$

$$R \approx (1.93)(0.119) - (5)(0.0141)$$

$$R \approx 0.229 - 0.035 = 0.194 	ext{ m}$$

Converting this to centimeters, we get $19.4 	ext{ cm}$, which is closest to **$20 	ext{ cm}$**. The beauty of tilting the axes is that it turns a daunting geometric nightmare into a straightforward plug-and-chug kinematic exercise!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Tilted Coordinate Axes$

$Resolving Initial Velocity$

$Resolving Acceleration$

$Time of Flight Formula$

$Calculating Time of Flight$

$Range Equation$

$Substituting Values for Range$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Projectile Motion

The Tilted World of Inclined Planes

Breaking Down the Vectors

The Time of Flight

The Final Range

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tilted Coordinate Axes

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resolving Initial Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resolving Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time of Flight Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Time of Flight

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Range Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for Range

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward