A plane is inclined at an angle α = 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) (a) 14 cm (b) 18 cm (c) 20 cm (d) 26 cm

Answer: (c). Range up an incline: R = (2u² sin θ cos (θ + α))/(gcos ²α) R = (2(4)sin 15° cos 45°)/(10 × 3/4) = (8(0.259)(0.707))/(7.5) R ≈ 0.195 m ≈ 20 cm So the answer is (c) 20 cm.

A plane is inclined at an angle α = 30° with respect to the… | JEE Main 2019 (10 Apr, Shift 2)

Q10JEE Main 2019 · 10 APR · SHIFT 2Difficult

Horizontal Projectile & Inclined Plane

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)

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Answer: (c)

Range up an incline: $R = \frac{2u^2 \sin\theta \cos(\theta + \alpha)}{g\cos^2\alpha}$
$R = \frac{2(4)\sin 15^\circ \cos 45^\circ}{10 \times \frac{3}{4}} = \frac{8(0.259)(0.707)}{7.5}$
$R \approx 0.195\text{ m} \approx 20\text{ cm}$
So the answer is (c) 20 cm.

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Horizontal Projectile & Inclined Plane question from JEE Main 2019 (10 Apr, Shift 2)

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