A large wedge BCD, having its inclined surface at an angle θ = 4 5 ∘ to the horizontal, is travelling horizontally leftwards with uniform velocity u = 10 m / s . At some instant a particle is projecte

Question

A large wedge BCD, having its inclined surface at an angle $θ = 4 5^{\circ}$ to the horizontal, is travelling horizontally leftwards with uniform velocity $u = 10 m / s$ . At some instant a particle is projected vertically up with speed $V = 20 m / s$ from point A on ground lying at some distance right to the lower edge B of the wedge. The particle strikes the incline BC normally w.r.t. incline, while it was falling $[g = 10 m / s^{2}]$ .

TestHub · Accepted Answer

y = 20 t − 2 1 ​ g t 2 = 20 ⋅ 3 − 5 ⋅ 9 = 15 m . x co n t a c t ​ − x B ​ = 15 . x = 0 : 0 = x B ​ + 15 ⇒ x B ​ ( t = 3 ) = − 15 m , ⟹distance from A is 15 m (B). x B ​ ( t ) = x B 0 ​ − 10 t , x B 0 ​ − 30 = − 15 ⇒ x B 0 ​ = 15 m . A B = 15 m ⟹(A). The wedge frame has a constant relative horizontal speed (no extra acceleration), so the trajectory remains parabolic (D) . Correct options: A, B, D.

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