A stone is dropped from the roof of a building. It is observed to pass a window of height 2.0 m in 0.2 s . (Take g = 10 m s ⁻² ). Consider the following statements: (I) The velocity of the stone at the top edge of the window is 9 m s ⁻¹ . (II) The velocity of the stone at the bottom edge of the window is 10 m s ⁻¹ . (III) The top edge of the window is 4.05 m below the roof. (IV) The stone takes 0.5 s to reach the top

Step-by-step solution

Let us evaluate each statement by analyzing the motion of the stone. For the motion across the window: Displacement s = 2.0 m Time t = 0.2 s Acceleration a = g = 10 m s ⁻² Using the second equation of motion to find the velocity at the top edge ( u ): s = ut + 1 2 at^2 2.0 = u(0.2) + 1 2 (10)(0.2)^2 2.0 = 0.2u + 5(0.04) 2.0 = 0.2u + 0.2 1.8 = 0.2u u = 9 m s ⁻¹ . Thus, statement (I) is correct. Velocity at the bottom edge of the window ( v ): v = u + at = 9 + 10(0.2) = 9 + 2 = 11 m s ⁻¹ . Thus, statement (II) is inc