Arubber ball dropped from a height of 81 feet and on each bounce it rebounds 48% of its previous height. how far has the ball traveled vertically at the moment it hits the ground for the 22nd time

Total distance traveled vertically by the ball at the 22nd time it hits ground is the sum of the  upward travel and the downward travel which equals to 230.54 feet. Both are geometric progressions.

Step-by-step explanation:

(1) Downward travel:

Initial height = a = 81 feet

On each impact to the ground, ball rebounds 48 % of its previous height, so:

Sum of 22 times it travels downward = a * [ { 1 - (0.48^22) } / ( 1 - 0.48 ) ]

                                                              = 81 * 1.92307

                                                              = 155.77 feet

(2) Upward (rebound) travel:

Initial height = a = 81 feet

Height after 1st rebound = b = 81 * 0.48 = 38.88 feet

Since the ball will rebound only 21 times, so we will calculate only distance of only 21 rebounds:

Sum of 21 times it travels upward = b * [ { 1 - (0.48^21) } / ( 1 - 0.48 ) ]

                                                        = 38.88 * 1.923

                                                        = 74.77 feet

So the total travel = 155.77 + 74.77 = 230.54 feet approximately

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x