These were my answers
a) (H0)*(kh)^n
b) (n+1)+ [E k=1 to infinity (H0)*(kh)^n]
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!
2 answers
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
3 answers