Calculus
posted by Anonymous on .
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 don't get how to solve this.

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 "upanddown" 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)/(4/51)  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 
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?