Asked by Aubrie
A ball rebounds 1/2 of the height from which it is dropped. It is dropped 56m from the building and it kept bouncing. How far would the ball have travelled up and down when it strikes the ground for the 6th time?
Answers
Answered by
Reiny
make a sketch of the first few bounces and you will see:
1 bounce --- 56 m
2nd bounce = 2(28) = 56
3rd bounce = 2(14) = 28
4th bounc = 2(7) = 14
5th bounce = 2(3.5) = 7
6th bounce = 2(1.75) = 3.5
add them up to get ..... 164.5
or:
from the 2nd on , we have a geometric sequence with
a = 56, r = 1/2
We need the sum of
1st term + 5 terms of the GS
= 56 + 56(1 - (1/2)^5)/(1 - 1/2)
= 56 + 56( (31/32) / (1/2) )
= 56 + 56(31/16)
= 56 + 217/2
= 164.5
1 bounce --- 56 m
2nd bounce = 2(28) = 56
3rd bounce = 2(14) = 28
4th bounc = 2(7) = 14
5th bounce = 2(3.5) = 7
6th bounce = 2(1.75) = 3.5
add them up to get ..... 164.5
or:
from the 2nd on , we have a geometric sequence with
a = 56, r = 1/2
We need the sum of
1st term + 5 terms of the GS
= 56 + 56(1 - (1/2)^5)/(1 - 1/2)
= 56 + 56( (31/32) / (1/2) )
= 56 + 56(31/16)
= 56 + 217/2
= 164.5
Answered by
Glaiza moros
Thank you
Answered by
Anonymous
Gak
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