A block is dropped from a height H₁ above a deep pool of liquid and falls freely under gravity. Upon hitting the liquid surface, a sudden drag reduces its velocity to 1 4 of its impact velocity. It then sinks at this new constant velocity to a depth H₂ . If the time spent moving in the liquid is 3 times the time spent falling in the air, the ratio H₁ H₂ is

Step-by-step solution

Let the impact velocity of the block just before hitting the liquid be v₁ . Using kinematics for free fall from rest, v₁ = 2gH₁ The time spent falling in the air is t₁ = 2H₁ g The velocity of the block in the liquid is constant and given as v₂ = v₁ 4 = 2gH₁ 4 The time spent in the liquid is t₂ = 3t₁ = 3 2H₁ g The depth H₂ covered in the liquid at constant velocity is H₂ = v₂ t₂ Substituting the expressions for v₂ and t₂ : H₂ = ( 2gH₁ 4 ) ( 3 2H₁ g ) H₂ = 3 4 2gH₁ 2H₁ g H₂ = 3 4 4H₁^2 = 3 4 2H₁ = 3 2 H₁ Rearranging