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 andĀ a distance of $$45*\frac{3}{5}$$ while going down again. after 2nd rebound it'll cover a distance ofĀ $$45*\frac{3}{5}*\frac{3}{5}$$Ā while going up andĀ a distance of $$45*\frac{3}{5}*\frac{3}{5}$$ while going down again and so onĀ
i.e. total distance traveled by the ball is 45 + $$45*\frac{3}{5}$$ +Ā $$45*\frac{3}{5}$$ +Ā $$45*\frac{3}{5}*\frac{3}{5}$$ +Ā $$45*\frac{3}{5}*\frac{3}{5}$$ +...........
this form 2 infinite GPs asĀ 45 + $$45*\frac{3}{5}$$ +Ā $$45*\frac{3}{5}*\frac{3}{5}$$+....... andĀ $$45*\frac{3}{5}$$+$$45*\frac{3}{5}*\frac{3}{5}$$+....
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*$$\frac{3}{5}$$Ā and r=$$\frac{3}{5}$$ therefore sum of this infinite GP isĀ $$\frac{45*\frac{3}{5}}{1-\frac{3}{5}}$$ = $$\frac{135}{2}$$
Therefore our answer =Ā $$\frac{225}{2}+\frac{135}{2}$$ = 180 m i.e. option 'B'
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