A plane is inclined at an angle α = 30 ° with respect to the horizontal. A particle is projected with a speed u = 2 m s - 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 m s - 2 )

Step-by-step solution

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 &#8