Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s . Ball B is thrown vertically down with an initial velocity u at t = 2 s . After a certain time, both balls meet 100 m above the ground. Find the value of u in m s - 1 . [use g = 10 m s - 2 ]

Step-by-step solution

Displacement covered by first ball, 180 - 100 = 0 + 1 2 × 10 × t 2 ⇒ t = 4   s Now, the second body gets only, t 2 = 4 - 2 = 2   s . Displacement covered by second will be same, 80 = u × 2 + 1 2 × 10 × 2 2 ⇒ u = 80 - 20 2 = 30   m   s - 1