On an inclined plane of inclination α =30° with respect to the horizontal, a particle is projected with a speed u=2ms- 1 from the base of the plane making an angle θ =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=10ms- 2 )

Question

On an inclined plane of inclination α =30° with respect to the horizontal, a particle is projected with a speed u=2ms- 1 from the base of the plane making an angle θ =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=10ms- 2 ) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-y428zx0qeueu8ihh.jpg" /> (A) 18cm (B) 14cm (C) 26cm (D) 20cm

Tardigrade Admin · Accepted Answer

On inclined plane (range) R=(2 U2 sin α cos (. α + β .)/g (cos)2 β ) Where α =15° ,β =30° ,U=2m/s On solving we get R=(4/5)((1/√3) - (1/3))≈20cm

$18 c m$

$14 c m$

$26 c m$

$20 c m$

On inclined plane (range) $R = \frac{2 U ^{2} s in α cos ( α + β )}{g ( cos ) ^{2} β}$
Where $α = 1 5^{\circ}, β = 3 0^{\circ}, U = 2 m / s$
On solving we get $R = \frac{4}{5} (\frac{1}{3} - \frac{1}{3}) \approx 20 c m$

18cm

14cm

26cm

20cm

On inclined plane (range) R=g(cos)2β2U2sinαcos(α+β)​ Where α=15∘,β=30∘,U=2m/s On solving we get R=54​(3​1​−31​)≈20cm

$18 c m$

$14 c m$

$26 c m$

$20 c m$

On inclined plane (range) $R = \frac{2 U ^{2} s in α cos ( α + β )}{g ( cos ) ^{2} β}$
Where $α = 1 5^{\circ}, β = 3 0^{\circ}, U = 2 m / s$
On solving we get $R = \frac{4}{5} (\frac{1}{3} - \frac{1}{3}) \approx 20 c m$