A ball is dropped vertically from height of 30 meters after bounces.The ball reaches a max height to 60% of height of the previous bounce. What are the terms of the sequence. What is the height to the nearest tenth of a meter of the ball after the tenth bounces?
h0 = 30
h1 = .6*30
h2 = .6^22 * 30
h3= .6^3*30
hn = .6^n*30
h10 = .6^10 * 30 = .2
This is a geometric progression
a1=30
r=.6
What is 30(.6)^10 ?
use the formula sn=(a1)(1-r^n)/(1-r).
a1=30
r=0.60
n=10
Viola :)
To find the height of the ball after each bounce, we can set up a geometric sequence where each term represents the maximum height reached after a bounce. The first term of the sequence is 30 meters, which is the initial height of the ball.
According to the given information, each term in the sequence is 60% of the previous term. This means the common ratio (r) of the geometric sequence is 0.6.
To find the terms of the sequence, we can use the formula for the nth term of a geometric sequence:
aₙ = a₁ * r^(n-1)
where aₙ is the nth term, a₁ is the first term, r is the common ratio, and n is the number of bounces.
Let's calculate the terms of the sequence up to the 10th bounce:
a₁ = 30 (initial height)
r = 0.6 (common ratio)
For n = 1:
a₁ = 30 * 0.6^(1-1) = 30 * 1 = 30 meters
For n = 2:
a₂ = 30 * 0.6^(2-1) = 30 * 0.6 = 18 meters
For n = 3:
a₃ = 30 * 0.6^(3-1) = 30 * (0.6 * 0.6) = 10.8 meters
Continuing this pattern, we can find the terms of the sequence up to the 10th bounce:
a₄ ≈ 6.5 meters
a₅ ≈ 3.9 meters
a₆ ≈ 2.3 meters
a₇ ≈ 1.4 meters
a₈ ≈ 0.9 meters
a₉ ≈ 0.5 meters
a₁₀ ≈ 0.3 meters
So, after the tenth bounce, the height of the ball is approximately 0.3 meters to the nearest tenth of a meter.