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=2(H)/3.After the 2nd bounce,the ball reaches a height of H_2 cm where H_2=2(H_1)/3.The ball continue bouncing in this way until it stops.
Given that H=600cm,find
b) the total distance, in cm,travelled by the ball until it stops.
@ganeshie8

Question

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=2(H)/3.After the 2nd bounce,the ball reaches a height of H_2 cm where H_2=2(H_1)/3.The ball continue bouncing in this way until it stops.
Given that H=600cm,find
b) the total distance, in cm,travelled by the ball until it stops.
@ganeshie8

ganeshie8 · Answer

H = 600cm = 6m

Plug that in and can you list the first few terms(heights of bounces) ?

anonymous · Answer

$$600,400,266\frac{ 2 }{ 3 },177\frac{ 7 }{ 9 },118\frac{ 14 }{ 27 },...$$

ganeshie8 · Answer

Do not simplify, leave 600 as it is in the terms

ganeshie8 · Answer

$H_0=600$

$H_1=600(2/3)$

$H_2=600(2/3)^2$

$H_3=600(2/3)^3$

$\vdots$

ganeshie8 · Answer

Those are the heights at each step, yes ?

anonymous · Answer

yes

ganeshie8 · Answer

Now, look at what the ball is doing

ganeshie8 · Answer

It is falling freely from a height of 600cm, so it must travel a distance of 600cm before hitting the ground for the first time, yes ?

anonymous · Answer

yes

ganeshie8 · Answer

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ganeshie8 · Answer

After that, it reaches a height of $H_1$ and also it falls by the same distance

ganeshie8 · Answer

If you see, except for $H_0$, all other heights are travelled "TWO" times. 
One while rising, other while falling.

ganeshie8 · Answer

remember the infinite geometric series formula ?

anonymous · Answer

yes

ganeshie8 · Answer

The distances travelled by the ball looks something like below : 


$D_0=600$

$D_1=\color{red}{2}*600(2/3)$

$D_2=\color{red}{2}*600(2/3)^2$

$D_3=\color{red}{2}*600(2/3)^3$

$\vdots$

ganeshie8 · Answer

Forget $D_0$.

Observe that the other terms form a geometric progression with $r = 2/3$

ganeshie8 · Answer

$D_1=\color{red}{2}*600(2/3)$

$D_2=\color{red}{2}*600(2/3)^2$

$D_3=\color{red}{2}*600(2/3)^3$

$\vdots$

ganeshie8 · Answer

first term = $\color{red}{2}*600(2/3)$

common ratio = $2/3$

can you find the infinite sum using the geometric series formula ?

anonymous · Answer

$$S_\infty=\frac{ a }{ 1-r }$$$$=\frac{ 800 }{ 1-\frac{ 2 }{ 3 } }$$

ganeshie8 · Answer

Yes, simplify

anonymous · Answer

$$=2400$$

ganeshie8 · Answer

Looks good. Add the initial height $H_0$ to that...

ganeshie8 · Answer

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