Question

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 ms^-1. [use g = 10 ms^-2] : 

(A) 10 

(B) 15 

(C) 20 

(D) 30

Answer 1

Correct option is (D) 30

Let they meet at time t.

Time taken by ball B to meet A = 2 sec

using S = ut + 1/2 at² –80

= –u x 2 + 1/2 (-10)(2)²

u = 30