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 $$v$$? (take $$g = 10\,m/{s^2}$$   )

Solution :
Clearly distance moved by 1^st ball in $$18\,s$$  = distance moved by 2^nd ball in $$12\,s.$$
Now, distance moved in $$18\,s$$  by 1^st ball $$ = \frac{1}{2} \times 10 \times {18^2} = 90 \times 18 = 1620\,\,m$$
Distance moved in $$12\,s$$  by 2^nd ball $$ = ut + \frac{1}{2}g{t^2}$$
$$\eqalign{ & \therefore 1620 = 12v + 5 \times 144 \cr & \Rightarrow v = 135 - 60 = 75\,m{s^{ - 1}} \cr} $$