A stone is projected with velocity 2 g h , so that it just clears two walls of equal height h , at distance of 2 h from each other. The time interval of passing between the two walls is

Step-by-step solution

2 h = u t u x = 2 h Δ t   ∵ u = u x , t = Δ t . . . . i h = u y t - 1 2 g t 2 g t 2 - 2 u y t + 2 h = 0 t 1 = 2 u y + 4 u y 2 - 8 g h 2 g t 2 = 2 u y - 4 u y 2 - 8 g h 2 g Δ t = t 1 - t 2 = 4 u y 2 - 8 g h g u y 2 = g 2 ( Δ t ) 2 4 + 2 g h u x 2 + u y 2 = u 2 = ( 2 g h ) 2 4 n 2 ( Δ t ) 2 + g 2 ( Δ t ) 2 4 + 2 g h = 4 g h g 2 4 Δ t 4 - 2 g h Δ t 2 + 4 h 2 = 0 Δ t = 2 g h ± 4 g 2 h 2 - 4 g 2 h 2 g 2 / 2 = 4 h g Δ t = 2 h g