Asked by Stuck in Math
When a certain type of ball falls froma height of h onto a hard level surface, it bounce straight back up to a height of kh, where k is constant with 0<k<1. suppose this ball is dropped from an initial height h0
a) let the terms of the sequence {hn}from n=1 to infinity represent the max height reached by the ball after its nth bounce on the floor. find a formula for the general term hn.
b) let the terms of the seq. {dn}from n=1 to infinity represent the distance traveled by the ball between the nth and (n+1)th bounces on the ground. find formula
I am so Lost with this pls help!
a) let the terms of the sequence {hn}from n=1 to infinity represent the max height reached by the ball after its nth bounce on the floor. find a formula for the general term hn.
b) let the terms of the seq. {dn}from n=1 to infinity represent the distance traveled by the ball between the nth and (n+1)th bounces on the ground. find formula
I am so Lost with this pls help!
Answers
Answered by
Stuck in Math
These were my answers
a) (H0)*(kh)^n
b) (n+1)+ [E k=1 to infinity (H0)*(kh)^n]
a) (H0)*(kh)^n
b) (n+1)+ [E k=1 to infinity (H0)*(kh)^n]
Answered by
Steve
No reason for H0 as well as h
the nth bounce height is just hn = h*k^n for n=0...
Note that the 0th term is just h*k^0 = h, the starting height.
The distance traveled between bounces is 2hn (a round trip up and down to hn)
So, dn = 2h*k^n
the nth bounce height is just hn = h*k^n for n=0...
Note that the 0th term is just h*k^0 = h, the starting height.
The distance traveled between bounces is 2hn (a round trip up and down to hn)
So, dn = 2h*k^n
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.