A plane is inclined at an angle $\alpha=30^{\circ}$ with respect to the horizontal. A particle is projected with a speed $u=2 \mathrm{~ms}^{-1}$, from the base of the plant, 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 $\mathrm{g}=10 \mathrm{~ms}^{2}$)

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Kinematics

A plane is inclined at an angle

α = 3 0^{\circ}

with respect to the horizontal. A particle is projected with a speed

u = 2 ms^{- 1}

, from the base of the plant, making an angle

θ = 1 5^{\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

g = 10 ms^{2}

)

Answer: (D)

Projected from the base at

θ = 1 5^{\circ}

to an incline tilted at

α = 3 0^{\circ}

, the particle lands back on the slope after travelling a distance along it of

R = \frac{2 u ^{2} s i n θ c o s ( θ + α )}{g c o s ^{2} α}

. Substituting

u = 2 m s^{- 1}

θ = 1 5^{\circ}

and

α = 3 0^{\circ}

R = \frac{2 \times 4 \times s i n 1 5 ^{\circ} \times c o s 4 5 ^{\circ}}{10 \times c o s ^{2} 3 0 ^{\circ}} = \frac{8 \times 0.259 \times 0.707}{10 \times 0.75} \approx 0.20 m

, close to

20 cm

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option