Suppose you drop a ball from a window 25 meters above the ground. the ball bounces to 75% 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? hint: r ≠ 1, where a1 is the first term and r is the common ratio

Begin by writing the formula; 

y=abˣ
y=25(0.75)ˣ

Now calculate vertical height for one way, y=25(0.75)ˣ using sum of geometric series 

S=a₁(1-rⁿ)/1-r 
S=25(1-0.75¹⁰)/(1-0.75) 
S=94.368 m (in one direction). 

Multiply by 2 since you also need to consider the distance travelled when going down; 94.3+94.3=188.737 m 

Since 25 is not initially accounted in this sum; 

188.737+25=213.737 m 

Hope I helped :) 

2160ft^3 i think you could use a formala to you out with this in the future