A plane is inclined at an angle ⍺ = 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 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 = 10 ms

Engineering

Physics

Projectile Motion

Question

A plane is inclined at an angle ⍺ = 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 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 = 10 ms^–2)

26 cm

20 cm

18 cm

14 cm

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$T = \frac{2 usinθ}{gcosα}$

$R = ucosθT - \frac{1}{2} {gsinαT}^{2}$

$= \frac{ucosθ 2 usinθ}{gcosα} - \frac{gsinα}{2} \frac{4 u^{2} \sin^{2} θ}{g^{2} \cos^{2} α}$

$= \frac{u^{2} \sin^{2} θ}{gcosα} - \frac{u^{2} sinα}{{gcos}^{2} α} {1 - \cos 2 θ}$

$= \frac{4 \times \frac{1}{2}}{10 \times \frac{\sqrt{3}}{2}} - \frac{u^{2} sinα}{{gcos}^{2} α} {1 - \frac{\sqrt{3}}{2}}$

$= \frac{4}{10 \sqrt{3}} - \frac{8}{30} {1 - \frac{\sqrt{3}}{2}}$

$= \frac{4}{5 \sqrt{3}} - \frac{8}{30} = \frac{8 \sqrt{3} - 8}{30} = \frac{8 (\sqrt{3} - 1)}{30} = 20 cm$