A ball is dropped from the top of the building. The ball takes 0.5 seconds to fall past the 3 m length of a window some distance from the top of the building. If the velocities of the ball at the top and at the bottom of the window are Vt and Vb respectively, then Vt + Vb = ? (Take g = 10 m.s^-2). (Round off the answer to the nearest integer)

Let the total height till bottom of the window be x m. Then, the height from the top of the building to the top of the window = (x - 3) m . Let time taken by the ball to reach from top of the building to the top of the window = t s, then the time taken to reach the bottom of the window from the top of the building is (t + 0.5) s. Applying the equation, (x - 3) = 5*t² x = 5 (t + 0.5)² Solving, we get t = 0.35 seconds x = 3.6125 m Applying the third equation of motion u = 0 m.s-1 Vt² - u² = 2*g*0.6125 Solving we get Vt = 3.5 m.s-1 Now again applying the third equation of motion, Vb² - u² = 2*g*3.6125 Solving we get Vb = 8.5 m.s-1 Adding the two Vt + Vb = 12 m.s-1 (Correct Answer)