Question

Help

Answer 1

with waht

Answer 2

\[Haziq~drops~a~ball~from~a~height~of~H~cm~above~the~floor.\]\[After~the~1st~bounce,the~ball~reaches~a~height~of~H_1~cm~where~H_1=\frac{ 2 }{ 3 }H.\]\[After~the~second~bounce,the~ball~reaches~a~height~of~H_2~cm~where~H_2=\frac{ 2 }{ 3 }H_1.\]\[The~ball~continue~bouncing~inn~this~way~until~\it~stops.\]

Answer 3

\[H_1=\frac{ 2 }{ 3 }H\]\[H_2=\frac{ 2 }{ 3 }H_1\]

Answer 4

Geometric sequence, if you second equation really says,
\(H_{2}=(2/3)H_1\)

Answer 5

Given that H=600cm,find
a) the number of bounces when the maximum height of the ball from the floor is less than 100cm for the first time.

Answer 6

You can think of it this way,

\(H_n\) - nth term
(although in higher maths it is used to denote something very abstruse)

\(r=2/3\)

Answer 7

And the first H (without subscripts, is (think of it as) \(H_0\)

Answer 8

\[T_n<100\]

Answer 9

i got \[n<4.419\]but the answer say it is\[n>4.419\]

Answer 10

So, you have a geom. sequence:

\(H_0\) (which is given to be 600)
\(H_1\quad =H_0\times (2/3) \)
\(H_2\quad =H_1\times (2/3) \quad or,=H_0\times (2/3)^2\)
\(H_3\quad =H_2\times (2/3) \quad or,=H_0\times (2/3)^3\)

and so forth, satisfying,
\(H_n\quad =H_{n-1} \times (2/3) \quad or,=H_0\times (2/3)^n\)

Answer 11

So you need to solve,
\(H_x<100\)
as I understand

Answer 12

But how come are you gettting not a whole number of bounces?

Answer 13

also your question is a bit wrong, because "for the first time" part is really confusing...

Answer 14

i know but i just don't understand why i get\[n<4.419\] but the answer show it is \[n>4.419\]

Answer 15

n is the number of bounces, so the more bounces there are the smaller the height is (and that is what you need - height smaller than 100).

So, the more bounces there are the smaller the height is. (Consiering the common ratio of 2/3 which is positive and less than 1).
Therefore if on bounce 4.419 you get 100cm, then you need more bounces to get less than 100 cm.

Answer 16

But, again, the number of bounces must be at least an integer.
This is same as dealing with \(3\pi^{\Large e}\) number of suits.

Answer 17

I guess they haven't made it precise enough, and just meant this as an example of showing algebraic tecniques.

Answer 18

This is my working...maybe u could check my working whether is it correct or not\[a=\frac{ 2 }{ 3 }(600)=400cm~and~r=\frac{ 2 }{ 3 }\]\[T_n<100\]\[400(\frac{ 2 }{ 3 })^{n-1}<100\]\[(\frac{ 2 }{ 3 })^{n-1}<\frac{ 1 }{ 4 }\]

Answer 19

\[(n-1)\log_{10}\frac{ 2 }{ 3 }<\log_{10}\frac{ 1 }{ 4 }\]\[n-1<3.419\]\[n<4.419\]

Answer 20

but the answer say it is\[n>4.419\]\[Therefore,n=5\]