Question

On an inclined plane of inclination $\alpha =30^\circ $ with respect to the horizontal, a particle is projected with a speed $u=2ms^{- 1}$ from the base of the plane making an angle $\theta =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 $g=10ms^{- 2}$ )

• A. $18cm$
• B. $14cm$
• C. $26cm$
• D. $20cm$

Answer 1

Answer: (D)

On inclined plane (range) $R=\frac{2 U^{2} sin \alpha cos \left(\right. \alpha + \beta \left.\right)}{g \left(cos\right)^{2} \beta }$
Where $\alpha =15^\circ ,\beta =30^\circ ,U=2m/s$
On solving we get $R=\frac{4}{5}\left(\frac{1}{\sqrt{3}} - \frac{1}{3}\right)\approx20cm$