A ball is projected vertically upward with an initial velocity of 50 m s - 1 at t = 0 s . At t = 2 s , another ball is projected vertically upward with same velocity. At t = ____ s , second ball will meet the first ball g = 10 m s - 2 .

Step-by-step solution

Distance covered by ball A is h = 50 2 - 1 2 × 10 × 2 2 = 100 - 20 = 80   m Now, at time t = 2   s , velocity of A is v A = 50 - 10 × 2 = 30   m   s - 1 Relative velocity of both balls after 2   s is v rel = 50 - 30 = 20   m   s - 1 Here, relative displacement x rel = 80   m   s - 1 and relative acceleration a rel = 0 . Time taken t = 80 20 = 4   s Time at which they will meet is t = 4 + 2 = 6   s