Question

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

Answer 1

hey here is your answer

Distance of 1st ball=
S= ut + 1/2gt2
= 0×t + 1/2 x 10 x 18×18
= 1620

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

I think my answer is capable to clear your confusion