Math Advanced Help (1 Viewer)

shellybellyjelly · Jun 5, 2021

Does anyone know how to prove the equations for c (i) ?? Thank you in advanced!

Yesterday, a tennis ball used in a game of cricket in the playground was hit onto the science block roof. Luckily it rolled off the roof. After bouncing on the playground it reached a height of 3 metres. After the next bounce it reached 2 metres, then 113 metres, and so on.

a What was T n , the height in metres reached after the nth bounce?
b What was the height of the roof that the ball fell from?
c The last time the ball bounced, its height was below 1 cm for the first time. After that it rolled away
across the playground.

Screen Shot 2021-06-05 at 10.45.39 am.png

CM_Tutor · Jun 5, 2021

shellybellyjelly said:
Does anyone know how to prove the equations for c (i) ?? Thank you in advanced!

Yesterday, a tennis ball used in a game of cricket in the playground was hit onto the science block roof. Luckily it rolled off the roof. After bouncing on the playground it reached a height of 3 metres. After the next bounce it reached 2 metres, then 113 metres, and so on.

a What was T n , the height in metres reached after the nth bounce?
b What was the height of the roof that the ball fell from?
c The last time the ball bounced, its height was below 1 cm for the first time. After that it rolled away
across the playground.
View attachment 30792

I am taking "2 metres, then 113 metres, and so on" to mean "2 metres, then $1\dfrac{1}{3}$ m, and so on."

The height of the bounces is a GP, with the first term (let's call it $h$ ) being the height of the roof. The GP is thus:

$h,\ h\left(\cfrac{2}{3}\right),\ h\left(\cfrac{2}{3}\right)^2,\ h\left(\cfrac{2}{3}\right)^3,\ ...$

$\begin{align*} \text{With the second term being $T_2 = 3\ \text{m}$, we have} \quad h\left(\cfrac{2}{3}\right) = 3 \quad &\implies \quad h = \cfrac{9}{2} = 4.5\ \text{m} \\ \text{which is confirmed from $T_3 = 2\ \text{m}$, where we also have} \quad h\left(\cfrac{2}{3}\right)^2 = 2 \quad &\implies \quad h = \cfrac{2 \times 3^2}{2^2} = \cfrac{3^2}{2} = 4.5\ \text{m} \\ \text{and also from $T_4 = 1\cfrac{1}{3} = \cfrac{4}{3}\ \text{m}$, where we have} \quad h\left(\cfrac{2}{3}\right)^3 = \cfrac{4}{3} \quad &\implies \quad h = \cfrac{2^2 \times 3^3}{3 \times 2^3} = \cfrac{3^2}{2} = 4.5\ \text{m} \end{align*}$

$\text{So, we have as our general rule} \quad T_n = h\left(\cfrac{2}{3}\right)^{n - 1} = \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{n - 1} \quad \text{for $n \in \mathbb{Z}^+$}$

Its final bounce occurs when $T_n < 1\ \text{cm}$ for the first time.

$\begin{align*} T_n &< 1\ \text{cm} = 0.01\ \text{m} \\ \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{n - 1} &< 0.01 \\ \left(\cfrac{2}{3}\right)^{n - 1} &< \cfrac{1}{100} \times \cfrac{2}{9} \\ \left(\cfrac{2}{3}\right)^{n - 1} &< \cfrac{1}{450} \\ \cfrac{1}{\cfrac{1}{450}} &< \cfrac{1}{\left(\cfrac{2}{3}\right)^{n - 1}} \quad \text{recalling that $0 < a < b < 1 \implies 1 < \cfrac{1}{b} < \cfrac{1}{a}$} \\ 1 \div \cfrac{1}{450} &< 1 \div \left(\cfrac{2}{3}\right)^{n - 1} \\ 1 \times \cfrac{450}{1} &< 1 \times \left(\cfrac{3}{2}\right)^{n - 1} \\ \left(\cfrac{3}{2}\right)^{n - 1} &> 450 \end{align*}$

I am not seeing why they have asked for a proof that the expression is greater than 300 when it is greater than 450 (unless I've made an error, which is always possible), but the result requested does automatically follow from the one above. However, I do see the reason for asking that the equation be rearranged to this form, as it avoids the need to recognise that $\ln{\dfrac{2}{3} < 0$ in the alterative working I'll put below. Anyway, carrying on:

$\begin{align*} \left(\cfrac{3}{2}\right)^{n - 1} &> 450 \\ \ln{\left(\cfrac{3}{2}\right)^{n - 1}} &> \ln{450} \\ (n - 1)\ln{\left(\cfrac{3}{2}\right)} &> \ln{450} \quad \text{as $\ln{a^b} = b\ln{a}$} \\ n - 1 &> \cfrac{\ln{450}}{\ln{\left(\cfrac{3}{2}\right)}} \\ n &> 1 + \cfrac{\ln{450}}{\ln{3} - \ln{2}} \quad \text{as $\ln{\cfrac{a}{b}} = \ln{a} - \ln{b}$} \\ n &> 16.0627... \end{align*}$

So, the final bounce occurs when $n = 17$ , and so there are 17 bounces.

Check:
$T_{16} = \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{16-1} = \cfrac{3^2}{2} \times \cfrac{2^{15}}{3^{15}} = \cfrac{2^{14}}{3^{13}} = \cfrac{16384}{1594323} = 0.010276...\ \text{m}$

$T_{17} = \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{17-1} = \cfrac{3^2}{2} \times \cfrac{2^{16}}{3^{16}} = \cfrac{2^{15}}{3^{14}} = \cfrac{32768}{4782969} = 0.0068509...\ \text{m}$

Alternative working:

$\begin{align*} T_n &< 1\ \text{cm} = 0.01\ \text{m} \\ \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{n - 1} &< 0.01 \\ \left(\cfrac{2}{3}\right)^{n - 1} &< \cfrac{1}{100} \times \cfrac{2}{9} \\ \ln{\left(\cfrac{2}{3}\right)^{n - 1}} &< \ln{\left(\cfrac{1}{450}\right)} \\ (n - 1)\ln{\left(\cfrac{2}{3}\right)} &< \ln{450^{-1}} \\ n - 1 &> \cfrac{-\ln{450}}{\ln{\left(\cfrac{2}{3}\right)}} \quad \text{as $\ln{\left(\cfrac{2}{3}\right)} < 0$} \\ n &> 1 - \cfrac{\ln{450}}{\ln{2}-\ln{3}} \\ n &> 1 + \cfrac{\ln{450}}{\ln{3}-\ln{2}} \\ n &> 16.0627... \end{align*}$

shellybellyjelly · Jun 6, 2021

S

CM_Tutor said:
I am taking "2 metres, then 113 metres, and so on" to mean "2 metres, then $1\dfrac{1}{3}$ m, and so on."

The height of the bounces is a GP, with the first term (let's call it $h$ ) being the height of the roof. The GP is thus:

$h,\ h\left(\cfrac{2}{3}\right),\ h\left(\cfrac{2}{3}\right)^2,\ h\left(\cfrac{2}{3}\right)^3,\ ...$

$\begin{align*} \text{With the second term being $T_2 = 3\ \text{m}$, we have} \quad h\left(\cfrac{2}{3}\right) = 3 \quad &\implies \quad h = \cfrac{9}{2} = 4.5\ \text{m} \\ \text{which is confirmed from $T_3 = 2\ \text{m}$, where we also have} \quad h\left(\cfrac{2}{3}\right)^2 = 2 \quad &\implies \quad h = \cfrac{2 \times 3^2}{2^2} = \cfrac{3^2}{2} = 4.5\ \text{m} \\ \text{and also from $T_4 = 1\cfrac{1}{3} = \cfrac{4}{3}\ \text{m}$, where we have} \quad h\left(\cfrac{2}{3}\right)^3 = \cfrac{4}{3} \quad &\implies \quad h = \cfrac{2^2 \times 3^3}{3 \times 2^3} = \cfrac{3^2}{2} = 4.5\ \text{m} \end{align*}$

$\text{So, we have as our general rule} \quad T_n = h\left(\cfrac{2}{3}\right)^{n - 1} = \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{n - 1} \quad \text{for $n \in \mathbb{Z}^+$}$

Its final bounce occurs when $T_n < 1\ \text{cm}$ for the first time.

$\begin{align*} T_n &< 1\ \text{cm} = 0.01\ \text{m} \\ \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{n - 1} &< 0.01 \\ \left(\cfrac{2}{3}\right)^{n - 1} &< \cfrac{1}{100} \times \cfrac{2}{9} \\ \left(\cfrac{2}{3}\right)^{n - 1} &< \cfrac{1}{450} \\ \cfrac{1}{\cfrac{1}{450}} &< \cfrac{1}{\left(\cfrac{2}{3}\right)^{n - 1}} \quad \text{recalling that $0 < a < b < 1 \implies 1 < \cfrac{1}{b} < \cfrac{1}{a}$} \\ 1 \div \cfrac{1}{450} &< 1 \div \left(\cfrac{2}{3}\right)^{n - 1} \\ 1 \times \cfrac{450}{1} &< 1 \times \left(\cfrac{3}{2}\right)^{n - 1} \\ \left(\cfrac{3}{2}\right)^{n - 1} &> 450 \end{align*}$

I am not seeing why they have asked for a proof that the expression is greater than 300 when it is greater than 450 (unless I've made an error, which is always possible), but the result requested does automatically follow from the one above. However, I do see the reason for asking that the equation be rearranged to this form, as it avoids the need to recognise that $\ln{\dfrac{2}{3} < 0$ in the alterative working I'll put below. Anyway, carrying on:

$\begin{align*} \left(\cfrac{3}{2}\right)^{n - 1} &> 450 \\ \ln{\left(\cfrac{3}{2}\right)^{n - 1}} &> \ln{450} \\ (n - 1)\ln{\left(\cfrac{3}{2}\right)} &> \ln{450} \quad \text{as $\ln{a^b} = b\ln{a}$} \\ n - 1 &> \cfrac{\ln{450}}{\ln{\left(\cfrac{3}{2}\right)}} \\ n &> 1 + \cfrac{\ln{450}}{\ln{3} - \ln{2}} \quad \text{as $\ln{\cfrac{a}{b}} = \ln{a} - \ln{b}$} \\ n &> 16.0627... \end{align*}$

So, the final bounce occurs when $n = 17$ , and so there are 17 bounces.

Check:
$T_{16} = \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{16-1} = \cfrac{3^2}{2} \times \cfrac{2^{15}}{3^{15}} = \cfrac{2^{14}}{3^{13}} = \cfrac{16384}{1594323} = 0.010276...\ \text{m}$

$T_{17} = \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{17-1} = \cfrac{3^2}{2} \times \cfrac{2^{16}}{3^{16}} = \cfrac{2^{15}}{3^{14}} = \cfrac{32768}{4782969} = 0.0068509...\ \text{m}$

Alternative working:

$\begin{align*} T_n &< 1\ \text{cm} = 0.01\ \text{m} \\ \cfrac{9}{2}\left(\cfrac{2}{3}\right)^{n - 1} &< 0.01 \\ \left(\cfrac{2}{3}\right)^{n - 1} &< \cfrac{1}{100} \times \cfrac{2}{9} \\ \ln{\left(\cfrac{2}{3}\right)^{n - 1}} &< \ln{\left(\cfrac{1}{450}\right)} \\ (n - 1)\ln{\left(\cfrac{2}{3}\right)} &< \ln{450^{-1}} \\ n - 1 &> \cfrac{-\ln{450}}{\ln{\left(\cfrac{2}{3}\right)}} \quad \text{as $\ln{\left(\cfrac{2}{3}\right)} < 0$} \\ n &> 1 - \cfrac{\ln{450}}{\ln{2}-\ln{3}} \\ n &> 1 + \cfrac{\ln{450}}{\ln{3}-\ln{2}} \\ n &> 16.0627... \end{align*}$

Sorry but I'm slightly confused. Where does the 9/2 come from for the general rule.

CM_Tutor · Jun 6, 2021

shellybellyjelly said:
S

Sorry but I'm slightly confused. Where does the 9/2 come from for the general rule.

$h = 4.5 = \cfrac{9}{2}$

shellybellyjelly · Jul 13, 2021

CM_Tutor said:
$h = 4.5 = \cfrac{9}{2}$

Thank You so much!

Math Advanced Help (1 Viewer)

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