A ping pong ball is dropped from a 45 metres high multi-storey building. The ball bounces back three fifth of the distance each time before coming to rest. The total distance traversed by the ball is:

When the ball is dropped from 45m height it will cover a distance of 45m while going down after rebound it'll cover a distance of $$45*\frac{3}{5}$$ while going up and a distance of $$45*\frac{3}{5}$$ while going down again. after 2nd rebound it'll cover a distance of $$45*\frac{3}{5}*\frac{3}{5}$$ while going up and a distance of $$45*\frac{3}{5}*\frac{3}{5}$$ while going down again and so on 

i.e. total distance traveled by the ball is 45 + $$45*\frac{3}{5}$$ + $$45*\frac{3}{5}$$ + $$45*\frac{3}{5}*\frac{3}{5}$$ + $$45*\frac{3}{5}*\frac{3}{5}$$ +...........

this form 2 infinite GPs as 45 + $$45*\frac{3}{5}$$ + $$45*\frac{3}{5}*\frac{3}{5}$$+....... and $$45*\frac{3}{5}$$+$$45*\frac{3}{5}*\frac{3}{5}$$+....

For 1st GP a=45 and r=$$\frac{3}{5}$$ therefore sum of this infinite GP is $$\frac{45}{1-\frac{3}{5}}$$ = $$\frac{225}{2}$$

For 2nd GP a=45*$$\frac{3}{5}$$  and r=$$\frac{3}{5}$$ therefore sum of this infinite GP is  $$\frac{45*\frac{3}{5}}{1-\frac{3}{5}}$$ = $$\frac{135}{2}$$

Therefore our answer = $$\frac{225}{2}+\frac{135}{2}$$ = 180 m i.e. option 'B'