A block of mass m is kept on a smooth wedge of height h at rest. At t = 0, wedge starts moving with constant acceleration $a = \frac{3}{2} g$ as shown in figure. V represents the relative velocity of block w.r.t. wedge at A & H is the maximum height achieved by the block measured from ground.

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In the wedge frame, the pseudo force ma drives the block up the 45° slope: a re l ​ = a cos 4 5 ∘ − g sin 4 5 ∘ = ( 2 3 g ​ − g ) 2 ​ 1 ​ = 2 2 ​ g ​ , L = sin 4 5 ∘ h ​ = h 2 ​ V = 2 a re l ​ L ​ = g h ​ Leaving A, the vertical component is v y ​ = V sin 4 5 ∘ , so it rises further Δ h = 2 g v y 2 ​ ​ = 4 g V 2 ​ = 4 h ​ : H = h + 4 h ​ = 4 5 h ​ Ans. (B,C)

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