Posted by **Anonymous** on Sunday, September 18, 2011 at 8:13pm.

Hard rubber ball is thrown down from a building with a height of 5 metres. After each bounce, the ball rises to 4/5 of its previous height. Total vertical distance d ball has travelled at the moment it hits the ground for the 8th time, to the nearest 10th of a metre, is..____________

I dont get how to solve this.

- Calculus -
**Reiny**, Sunday, September 18, 2011 at 9:02pm
make a sketch and you will see that the total distances

= 5 + 2(5)(4/5) + 2(5)(4/5)^2 + .... for 8 terms

= 5 + 10(4/5) + 10(4/5)^2 + ... for 8 terms

notice that the first term is not part of the pattern since it does not complete the "up-and-down" path

lets make it fit the patters:

total distance

= (10 + 10(4/5) + 10(4/5)^2 + ... 10(4/5)^7) - 5

= 10 ((4/5)^8 - 1)/(4/5-1) - 5

= 36.6

check:

1st bounce = 5

2nd bounce = 8 , 4 up and 4 down

3rd bounce = 6.4

3rd bounce = 5.12

4th bounce = 4.096

5th bounce = 3.2768

6th bounce = 2.62144

7th bounce = 2.097152

8th bounce = 1.6777216

sum of those 8 terms = 36.611392

- Calculus -
**izzeldin**, Wednesday, September 21, 2011 at 6:24am
A ball with mass 0.15 kg is thrown upward with initial velocity 20 m per sec from a roof of a building 30 m high find the max. height a ball reach?

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