Question

Suppose you drop a ball from a window 10 meters above the ground. The ball bounces to 80% of its previous height with each bounce. What is the total number of meters (rounded to the nearest tenth) the ball travels between the time it is dropped and the 10th bounce?

79.3 meters
73.2 meters
89.3 meters
77.5 meters

Answer 1

Total height = 10 m

Drops 10 m
1st bounce = 0.8*10 = 8 m up
+ 8 m down
2nd bounce = 0.8*8 = 6.4 m up
+ 6.4 m down
3rd bounce = 0.8*6.4 = 5.12 m up
+ 5.12 m down
....
9th bounce = 10(0.8)^9 = 1.34 m up
+ 1.34 m down

See the pattern? each bounce multiplies another 0.8 ; count the distance twice for returning to the ground
10, 8, 6.4, 5.12, ... 1.34
term in sequence = 10(0.8)^(n-1)
where n = {1,2,3,...10}
Total meters traveled
= ∑ 10(0.8)^0 + 2[10(0.8)^1 + 10(0.8)^2 + ..... 10(0.8)^9]
multiply sum by 2 then subtract the extra 10 m

Sum of a geometric sequence for the bounces
where n in (n-1)
`S = a( (1- r^(n) )/(1-r) ) \\ \\ S = 10 ( (1- 0.8^(10) )/(1-0.8) ) \\ \\ S = 44.63 \\ \\ 2*44.63 = 89.26 \\ \\ 89.26 - 10 = 79.3 m`would be 10 for the number of bounces before 10th bounce, "a" is 10.