A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h . Find the ratio of the times in which it is at height h 3 while going up and coming down respectively.

Step-by-step solution

The maximum height of a projectile thrown vertically upward is given by, h = u 2 2 g . ⇒ u = 2 g h Using equation of motion with constant acceleration, s = u t + 1 2 a t 2 ⇒ h 3 = 2 g h t - 1 2 g t 2 ⇒ g t 2 2 - 2 g h t + h 3 = 0 Solving the above quadratic equation we will get the times t 1 and t 2 . ⇒ t 2 t 1 = 2 g h + 2 g h - 4 × g 2 × h 3 2 g h - 2 g h - 4 × g 2 × h 3 = 2 g h + 4 g h 3 2 g h - 4 g h 3 = 3 + 2 3 - 2 Thus, the ratio t 1 t 2 = 3 - 2 3 + 2