Posted by **ALI** on Wednesday, October 3, 2012 at 1:42am.

A ball bounces two-thirds of the distance it falls. If it is dropped from a height of 10 meters, how far does it move before hitting the floor for the fourth time?

- Maths -
**Reiny**, Wednesday, October 3, 2012 at 8:15am
Distance traveled :

first bounce -- 10 m

2nd bounce -- (2/3)(10) or 20/3 both up and down

3rd bounce -- (2/3)(20/3) or 40/9 both up and down

4th bounce -- (2/3)(40/9) or 80/27 both up and down

total distance

= 10 + 2(20/3) + 2(40/9) + 2(80/27)

= 886/27 m

or appr 32.8 m

- Maths -
**Anonymous**, Monday, October 8, 2012 at 1:20am
hi Reiny,

U solved absolutely wrong, plz Ali follow this method

Solution:

For Downward Distance:-

a=10 (first term)

r=2/3 (common ratio)

n=4 (no. of terms)

s=? (sum of 4th terms)

Sn= a(1-r^n)/1-r (because r is less than1)

S= 10(1-(2/3)^4/1-(2/3)= 650/27

For Upward Distance

a=10*2/3=>20/3

r=2/3

n=3

S=?

S=a(1-r^n)/1-r

S=20/3(1-(2/3)^4/1-(2/3)

S=380/27

Now, add downward distance & upward distance for calculating total distance

650/27 + 380/27 = 1030/27

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