Posted by Angie on Friday, June 8, 2012 at 5:15pm.
1. 2↓,
2. 1.8↑, 1.8↓,
3. 1.62↑,1.62↓.
4. 1,46↑, 1.46↓,
5. 1.31↑, 1.31↓,
6. 1.18↑ , 1.18↓,
7. 1.07↑, 1.07↓,
8. 0.96↑,
(a) Σ =19.84 m.
(b) 0.96 m
the question is asking for the eighth time it hits the ground so wouldn't that include the other part of the .96, because what you have here is technically hitting the ground 7 times and the ball is in mid-air.... and for b) how did you set the problem up?
Thank you!
Angie's right. The ball needs to fall down after the 7th bounce, i.e. we stop counting after the 8th down arrow, to the sum 19.84, we need to add 0.96.
This is actually a summation of a geometric series of common ratio r=0.9, and initial value of 2.0.
The total (up and down) distance is twice the sum less 2 because the ball did not bounce for the first time.
The sum S(n) of the geometric series is
S(n)=a(1-r^n)/(1-r)
So for n=8,
S(8)=2(1-0.9^8)/(1-0.9)
=11.39066
So the total distance travelled is twice the sum less 2 (initial up motion)
=2*S(8)-2
=20.78
For part (b), the height h of the ball after the nth bounce is ar^n, where a=initial height, r=rebound factor, and n=number of bounces.
For a=2, r=0.9, n=8,
h=0.9^8=0.86
Thank you for the clarification on A and for B the way the question is worded I think I'm going to go with the .96...
THANKS GUYS!
You're welcome!
Note:
If it is a trick question, the height of the ball on the eighth bounce is zero!