Question

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

Answer 1

нєу тнєяє ιѕ αиѕωєя !!
ι нσρє уσυ нєℓρ !!!

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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

Answer 2

$\huge\mathbb\green{Answer}$

Clearly distance moved by 1st ball in 18 s = distance moved by 2nd ball in 12 s.

Now, distance moved in 18 s by 1st ball 1/2 x 10 x 18²

= 90 x 18 = 1620 m.

Distance moved in 12 s by 2nd ball
= ut + 1/2 gt²

1620 = 12v + 5 x 144

v = 135 – 60 = 75 ms