A marble is dropped from a height of 3 metres onto the ground. After hitting the ground, it bounces and reaches 80% of the height from which it was dropped. This repeats multiple times. Each time it bounces, the marble reaches 80% of the height previously reached. Eventually, the marble comes to rest on the ground.

What is the maximum distance that the marble travels from the time it was dropped until it comes to rest?

Given the ball falls from a height of 3 meters.

The ball reaches a height which 0.8 times the original height every time.

Hence this is in the form of a geometric progression. We need to count distance when the ball flies upward and downward.

Hence considering every time the ball flies upward to a series with terms :

h1, h2,..........................

Every time the ball falls down to be

d1, d2 ,...............

h1 = (0.8)*3, h2 = (0.8)*(0.8)*3 ,.........................

d1 = 3, d2 =3*(0.8), d3 = 3*(0.8)*(0.8)................

h1+h2 ........ = Sum of an infinite geometric progression. = 3*0.8(1+0.8+$$0.64+...............$$)

THe sum of an infinite GP with r less than 1 is : $$\frac{a}{1-r}$$

= $$2.4\cdot\left(\frac{1}{1-0.8}\right)\ =\ 12\ meters$$

The sum of d1+d2++........................

= 3 + (h1+h2+..................) = 15.

The total distance = 15 + 12 = 27 meters.