When a ball is dropped into a lake from a height 4 . 9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v . It reaches the bottom of the lake 4 . 0 s after it is dropped. The approximate depth of the lake is

Step-by-step solution

Using s = u t + 1 2 a t 2 Initially, u = 0 , so time taken to reach from A to B is t = 2 s g = 2 × 4 . 9 9 . 8 = 1   s Now, using v 2 = u 2 + 2 a s , we have v = 2 gh = 2 × 9 . 8 × 4 . 9 = 9 . 8   m   s - 1 ∵   t 2 = 4 - 1   s = 3   s ∴     d = t 2 × v = 3 × 9 . 8 = 29 . 4   m