Question

A plane is inclined at an angle $\alpha$ = 30° with respect to the horizontal. A particle is projected with a speed u = 2 $ms^{-1}$, 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 $ms^{-2}$]

• A. 0.20 m
• B. 0.25 m
• C. 0.27 m
• D. 0.30 m

Answer 1

Answer: (C)

0.27 m

Answer 2

The range $R$ of a projectile on an inclined plane, when projected with initial speed $u$ at an angle $\theta$ with respect to the plane, is given by the formula: $R = \frac{u^2 \sin(2\theta)}{g \cos^2 \alpha}$

Given: $u = 2 \, ms^{-1}$ $\alpha = 30^\circ$ $\theta = 15^\circ$ $g = 10 \, ms^{-2}$

Substitute the values into the formula: $R = \frac{(2 \, ms^{-1})^2 \sin(2 \times 15^\circ)}{10 \, ms^{-2} \times \cos^2(30^\circ)}$ $R = \frac{4 \sin(30^\circ)}{10 \times (\frac{\sqrt{3}}{2})^2}$ $R = \frac{4 \times \frac{1}{2}}{10 \times \frac{3}{4}}$ $R = \frac{2}{\frac{30}{4}}$ $R = \frac{2 \times 4}{30}$ $R = \frac{8}{30} = \frac{4}{15} \, m$

Calculating the numerical value: $R \approx 0.2667 \, m$

The distance is close to 0.27 m.