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!
These were my answers
a) (H0)*(kh)^n
b) (n+1)+ [E k=1 to infinity (H0)*(kh)^n]
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
To solve this problem, let's break it down step by step.
a) To find a formula for the general term hn, we need to determine how the height changes after each bounce. We know that the ball initially falls from a height of h₀ and bounces back up to a height of kh. For subsequent bounces, the ball will fall from the previous height and bounce back up to a height of kh again.
Let's consider the first few bounces:
1st bounce: h₁ = kh₀
2nd bounce: h₂ = kh₁ = k(kh₀) = k²h₀
3rd bounce: h₃ = kh₂ = k(k²h₀) = k³h₀
We can observe that after each bounce, the height is multiplied by the constant k. So, after the nth bounce, the height would be hn = kⁿh₀.
Therefore, the formula for the general term hn is hn = kⁿh₀.
b) To find a formula for the distance traveled by the ball between the nth and (n+1)th bounces on the ground, we need to consider that the ball travels down and then back up during each bounce.
Let's consider the first few distances:
Between the 1st and 2nd bounce, the ball travels down from an initial height of h₀ to the ground, and then back up to a height of h₁. So, the distance traveled is d₁ = h₀ + h₁.
Between the 2nd and 3rd bounce, the ball travels down from a height of h₁ to the ground, and then back up to a height of h₂. The distance traveled is d₂ = h₁ + h₂.
Similarly, we can observe a pattern:
d₁ = h₀ + h₁
d₂ = h₁ + h₂
d₃ = h₂ + h₃
From the previous part, we know the formula for hn is hn = kⁿh₀. Using this formula, we can rewrite the distances as:
d₁ = h₀ + k²h₀
d₂ = k²h₀ + k³h₀
d₃ = k³h₀ + k⁴h₀
We can simplify these expressions:
d₁ = (1 + k²)h₀
d₂ = (k² + k³)h₀
d₃ = (k³ + k⁴)h₀
We can see that the pattern appears to be that between each bounce, the distance traveled is the sum of the initial height and the previous height. So, we can conclude that the formula for the distance traveled between the nth and (n+1)th bounces is:
dn = (kⁿ + kⁿ⁺¹)h₀.
Therefore, the formula for the general term dn is dn = (kⁿ + kⁿ⁺¹)h₀.