A ball is dropped from a high-rise platform at t=0 starting from rest. After 6 s, 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?

The distance traveled by the first ball in t=18s is
h1 = ut + (1/2)gt² = 0(18) + 0.5(10)(18)² = 1620 m

As the second ball travels the same distance h1 in time t = (18 - 6) = 12 s, so
1620 = vt + 0.5gt²
or 1620 = 12v + 0.5(10)(12)²
or 12v = 1620 - 720
or 12v = 900
So, v = 75 m/s