A plane is inclined at an angle α = 30° with a respect to the horizontal. A particle is projected with a speed u = 2 ms-1 from the base of the plane, making an angle q = 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)

Question

A plane is inclined at an angle α = 30° with a respect to the horizontal. A particle is projected with a speed u = 2 ms-1 from the base of the plane, making an angle q = 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) 20 cm (C) 18 cm (D) 26 cm

Tardigrade Admin · Accepted Answer

t = (2×2× sin 15°/g cos 30°) S= 2 cos 15° × t - (1/2) g sin30° t2 Put values and solve S ≃ 20cm

14 cm

20 cm

18 cm

26 cm

t=gcos30∘2×2×sin15∘​ S=2cos15∘×t−21​gsin30∘t2 Put values and solve S ≃ 20cm

$t = \frac{2 \times 2 \times s i n 1 5 ^{\circ}}{g c o s 3 0 ^{\circ}}$
$S = 2 cos 1 5^{\circ} \times t - \frac{1}{2} g sin 3 0^{\circ} t^{2}$
Put values and solve
S $≃$ 20cm