A plane is inclined at an angle $ \alpha=30^{\circ} $ with respect to the horizontal. A particle is projected with a speed $ \mathrm{u}=2 \mathrm{~m} $ s $ ^{-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 $ \mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2} $ )

Question

A plane is inclined at an angle $ \alpha=30^{\circ} $ with respect to the horizontal. A particle is projected with a speed $ \mathrm{u}=2 \mathrm{~m} $ s $ ^{-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 $ \mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2} $ ), $ 20 \mathrm{~cm} $, $ 18 \mathrm{~cm} $, $ 14 \mathrm{~cm} $, $ 26 \mathrm{~cm} $

MARKS App · Accepted Answer

Range = 2 u 2 s i n α cos ⁡ α + θ g cos 2 ⁡ θ α = a n g l e o f i n c l i n a t i o n = 30 ° θ = a n g l e o f p r o j e c t i o n f r o m i n c l i n a t i o n = 15 ° u = 2 m / s R = 2 2 2 s i n 15 ° cos ⁡ 15 ° + 30 ° g cos 2 ⁡ 30 ° = 8 sin ⁡ 15 ° c o s 45 ° g c o s 2 30 ° s i n 15 ° = 3 - 1 2 2 R = 4 5 3 - 1 3 ≈ 0.2 m R ≈ 20 c m

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