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?

Answer 1

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

Answer 2

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

Answer 3

You are Right