Question

A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18 s.What is the value of v ?

Answer 1

Distance of 1st ball= S= 1/2 x 10 x 18 
= 1620

Distance of 2nd ball = S' = ut + 1/2gt²
1620 = 12u + 5 x 144
u = 135 - 60
= 75 m/s

Answer 2

the vertical distance traveled by ball 1 by t = 18 sec
  = s1 = u * t + 1/2 g t²
=> s1 = 0 + 1/2 g * 18² = 162 g

distance traveled by ball 2 in 12 secs. from t=6 sec to 18 sec. 
 = s2 = v * 12 + 1/2 g *12² = 12 v  + 72 g

s1 = s2
=>   12 v = 90 g
=>   v = 7.5 g or  75 m/sec