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Help !! Maths c2 question on series and sequence watch

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    HEY :P

    I was doing this question
    When a ball is dropped onto a horizontal floor it bounces such that it reaches a maximum height
    of 60% of the height from which it was dropped.

    a )Find the maximum height the ball reaches after its fourth bounce when it is initially dropped from 3 metres above the floor.

    after 4th bounce,
    reaches 3 × (0.6)^4
    = 0.3888 m

    im confused,is it arithmetic or geometric ,when i tried it i assumed it would be geometric but then the power would be to 3 so im wrong

    b) Show that when the ball is dropped from a height of h metres above the floor it travels a total distance of 4h metres before coming to rest.

    total distance
    = h + 2[0.6h + (0.6)^2
    h + (0.6)^3
    h + …]
    = h + 2 × S∞ of GP, a = 0.6h, r = 0.6
    = h + 2 0.6
    1 0.6
    × h

    = h + 3h = 4h metres

    could someone explain the second part to me as well

    Thank you ,
    i would really appreciate the help as im really confused on this question
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    (Original post by Sadilla)
    HEY :P

    I was doing this question
    When a ball is dropped onto a horizontal floor it bounces such that it reaches a maximum height
    of 60% of the height from which it was dropped.

    a )Find the maximum height the ball reaches after its fourth bounce when it is initially dropped from 3 metres above the floor.

    after 4th bounce,
    reaches 3 × (0.6)^4
    = 0.3888 m

    im confused,is it arithmetic or geometric ,when i tried it i assumed it would be geometric but then the power would be to 3 so im wrong
    This is fine, it is indeed a geometric progression for the sequence of heights.

    After 0 bounces: 3
    After 1 bounce: 3\cdot 0.6
    After 2 bounces: 3 \cdot 0.6^2
    ...
    After n bounces: 3 \cdot 0.6^n

    b) Show that when the ball is dropped from a height of h metres above the floor it travels a total distance of 4h metres before coming to rest.

    total distance
    = h + 2[0.6h + (0.6)^2
    h + (0.6)^3
    h + …]
    = h + 2 × S∞ of GP, a = 0.6h, r = 0.6
    = h + 2 0.6
    1 0.6
    × h

    = h + 3h = 4h metres

    could someone explain the second part to me as well

    Thank you ,
    i would really appreciate the help as im really confused on this question
    When a ball is dropped, it goes down a distance of h metres. So we have T=h so far as the total distance. When it bounces, it comes back up to a height 0.6h before travelling down to the ground a total distance of 0.6h as well. So the total distance now is T=h+0.6h+0.6h = h+2h(0.6). Similarly, after the second bounce, it will travel a distance of 2\cdot (0.6^2h) hence we have T=h+2h(0.6+0.6^2) and so on.

    We are looking at what we get when the ball becomes at rest, ie when we sum all these bounces which yields an infinite sum.

    We get: h+2h(0.6+0.6^2+0.6^3+...) which we can calculate by the infinite GP sum formula.
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    [QUOTE=RDKGames;76562278]This is fine, it is indeed a geometric progression for the sequence of heights.

    After 0 bounces: 3
    After 1 bounce: 3\cdot 0.6
    After 2 bounces: 0.3 \cdot 0.6^2
    ...
    After n bounces: 0.3 \cdot 0.6^n

    thank you , you said for the 2 bounces it 0.3 x(0.6)^2 but then how comes the 1st bounce was 3x0=0.6 ? so how comes its 3 and then 0.3 ?
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    (Original post by Sadilla)

    thank you , you said for the 2 bounces it 0.3 x(0.6)^2 but then how comes the 1st bounce was 3x0=0.6 ? so how comes its 3 and then 0.3 ?
    Ooops, sorry, got derailed. Fixed now.
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    (Original post by RDKGames)
    Ooops, sorry, got derailed. Fixed now.
    its okay ;p , i was also confused as to why the it goes back down the second time its still 2h ?
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    (Original post by Sadilla)
    its okay ;p , i was also confused as to why the it goes back down the second time its still 2h ?
    What are you referring to?
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    (Original post by RDKGames)
    What are you referring to?
    when you said before travelling back down and has height 0.6h , how comes it not times by 0.6 as its 60% of that height ?
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    (Original post by Sadilla)
    when you said before travelling back down and has height 0.6h , how comes it not times by 0.6 as its 60% of that height ?
    Because it hasn't bounced yet...

    A ball hit the ground, bounces back up to a height of 0.6h then it falls down the same amount of 0.6h before hitting the ground again and bouncing. Get it?
 
 
 
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