Consider a ball that is dropped from a height of 96ft in an absolute vacuum under the influence of Earth's gravitational field (the acceleration due to which is 32 ft/s^2). If the ball always bounces back to 2/3 of the height it had reached on the previous bounce then...
how far does tha ball travel before it stops bouncing??
how far does tha ball travel before it stops bouncing??
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I think U should define when ball stops bouncing - because they way this is stated, the ball will theoretically NEVER stop bouncing.
Of course there is a practical limit that could be stated as n = ?
where n = bounce number and the distance of bounce which is considered negligible.
The bounce distance starts with 96 ft after n = 1;
then it becomes => (192)(2/3)^(n+1) where the term (192)(2/3)^(n-1)
represents the total distance for a given bounce number.= n {after the 1st bounce}.
For n= 2 {2nd bounce) the distance of THIS bounce = 192(2/3)^1 = 128 ft.
For n= 3 the distance of THIS bounce {fallen+risen distance}= 192(2/3)^2 = 85.333 ft
For n= 4 the distance of THIS bounce {fallen+risen distance}= 192(2/3)^3 = 56.888 ft
after 25 bounces, we might consider that the ball has "stopped bouncing" as when
n = 25, the total distance of the twenty fifth bounce = 192(2/3)^25 = 7.60E-3 ft < 0.1"
so perhaps after 25 bounces the total distance may be summed to find the answer
for this infinite series?
{this series does have a limit so I guess that limit would be the answer
but I like the idea of stating a bounce no. that is considered to be the
last bounce.. which is more practical than mathematical :>)
Of course there is a practical limit that could be stated as n = ?
where n = bounce number and the distance of bounce which is considered negligible.
The bounce distance starts with 96 ft after n = 1;
then it becomes => (192)(2/3)^(n+1) where the term (192)(2/3)^(n-1)
represents the total distance for a given bounce number.= n {after the 1st bounce}.
For n= 2 {2nd bounce) the distance of THIS bounce = 192(2/3)^1 = 128 ft.
For n= 3 the distance of THIS bounce {fallen+risen distance}= 192(2/3)^2 = 85.333 ft
For n= 4 the distance of THIS bounce {fallen+risen distance}= 192(2/3)^3 = 56.888 ft
after 25 bounces, we might consider that the ball has "stopped bouncing" as when
n = 25, the total distance of the twenty fifth bounce = 192(2/3)^25 = 7.60E-3 ft < 0.1"
so perhaps after 25 bounces the total distance may be summed to find the answer
for this infinite series?
{this series does have a limit so I guess that limit would be the answer
but I like the idea of stating a bounce no. that is considered to be the
last bounce.. which is more practical than mathematical :>)