A rubber ball dropped on a hard surface takes a sequence of bounces, each one 4/5 as high as the preceding one. If this ball is dropped from a height of 10 feet, what is the total vertical distance it has traveled at the time when it hits the surface for its fifth bounce? A) 28 77/125 feetB) 49 1/25 feetC) 57 29/125 feetD) 59 29/125 feet

Answer:The answer is 57 29/125 feet ⇒ answer (C)Step-by-step explanation:∵ The ball dropped from a height 10 feet∴ The vertical distance = 10 feet∵ It bounces 4/5 as high as the preceding one∴ 1st bounce = 10 × 4/5 = 8 (move down another 8) ∴ The vertical distance = 8 + 8 = 16 feet ∴ 2nd bounce = 8 × 4/5 = 6.4 (move down another 6.4) ∴ The vertical distance = 6.4 + 6.4 = 12.8 feet ∴ 3rd bounce = 6.4 × 4/5 = 5.12 (move down another 5.12) ∴ The vertical distance = 5.12 + 5.12 = 10.24 feet ∴ 4th bounce = 5.12 × 4/5 = 4.096 (move down another 4.096) ∴ The vertical distance = 4.096 + 4.096 = 8.192 feet ∵ The ball will hit the surface to make the 5th bounce∴ The total vertical distance is :    10 + 16 + 12.8 + 10.24 + 8.192 = 57.232 = 57 29/125Another way:* We can use the formula of geometric sequence with   1st bounce distance (a) = 8 + 8 = 16 and ratio (r) = 4/5   and number of bounces (n) = 4 (hit the surface for its 5th bounce)∵ [tex]S_{n}=\frac{a(1-(r^{n}))}{1-r}[/tex]∴ [tex]S_{4}=\frac{16(1-(\frac{4}{5})^{4})}{1-\frac{4}{5}}=47\frac{29}{125}[/tex]* Then we must to add the 10 feet   (the ball dropped from height 10 feet)∴ The total vertical distance = 10 + 47 29/125 = 57 29/125 feet