A plane is inclined at an angle $$\alpha = 30$$° with respect to the horizontal. A particle is projected with a speed u = 2 m s$$^{-1}$$, from the base of the plane, making an angle $$\theta = 15$$° 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 = $$10\ m \ s^{-2}$$)

To find the distance from the base at which the particle hits the inclined plane (the range along the incline, $$R$$), we can break down the motion parallel and perpendicular to the inclined surface.

Here is the step-by-step formatting and solution of your approach.

1. Given Data

• Angle of inclination of the plane: $$\alpha = 30^\circ$$
• Angle of projection relative to the plane: $$\theta = 15^\circ$$
• Initial velocity: $$u = 2\text{ m s}^{-1}$$
• Acceleration due to gravity: $$g = 10\text{ m s}^{-2}$$
• $$u_x = u \cos\theta = 2 \cos 15^\circ$$
• $$u_y = u \sin\theta = 2 \sin 15^\circ$$
• $$a_x = -g \sin\alpha = -10 \sin 30^\circ$$
• $$a_y = -g \cos\alpha = -10 \cos 30^\circ$$

2. Components of Motion

By choosing the $$x$$-axis along the inclined plane (upwards) and the $$y$$-axis perpendicular to the inclined plane:

Initial Velocity Components

Acceleration Components

3. Calculation of Time of Flight ($$T$$)

The particle returns to the inclined plane when its displacement perpendicular to the plane is zero ($$y = 0$$):

$$y = u_y T + \frac{1}{2} a_y T^2 = 0$$

$$(2 \sin 15^\circ) T - \frac{1}{2} (g \cos 30^\circ) T^2 = 0$$

Solving for non-zero time $$T$$:

$$T = \frac{2 u \sin\theta}{g \cos\alpha} = \frac{2 \times 2 \sin 15^\circ}{10 \cos 30^\circ}$$

4. Range Along the Incline ($$R$$)

The displacement along the incline is given by:

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

Substituting $$u_x = 2 \cos 15^\circ$$ and $$a_x = -g \sin 30^\circ$$:

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

Alternative Standard Formula

Instead of plugging numbers into both terms manually, we can use the simplified standard formula for the range up an inclined plane:

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

Substituting the values into this formula makes it much cleaner:

$$R = \frac{2 \times (2)^2 \times \sin 15^\circ \times \cos(15^\circ + 30^\circ)}{10 \times \cos^2 30^\circ}$$

$$R = \frac{8 \times \sin 15^\circ \times \cos 45^\circ}{10 \times \left(\frac{\sqrt{3}}{2}\right)^2}$$

$$R = \frac{8 \times \sin 15^\circ \times \frac{1}{\sqrt{2}}}{10 \times \frac{3}{4}} = \frac{32 \sin 15^\circ}{30\sqrt{2}}$$

Using the known value $$\sin 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}}$$:

$$R = \frac{32 \times (\sqrt{3} - 1)}{30 \times 2 \times \sqrt{2} \times \sqrt{2}} = \frac{32(\sqrt{3}-1)}{120} = \frac{4(\sqrt{3}-1)}{15}\text{ meters}$$

5. Final Numeric Approximation

Using $$\sqrt{3} \approx 1.732$$:

$$R \approx \frac{4 \times (1.732 - 1)}{15} = \frac{4 \times 0.732}{15} = \frac{2.928}{15} \approx 0.1952\text{ m}$$

Converting meters to centimeters:

$$R \approx 0.1952 \times 100\text{ cm} = 19.52\text{ cm} \approx 20\text{ cm}$$

Correct Answer: A (20 cm)