Alan

The last one is correct.

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1.  Use s = ut + (1/2)at², where s = distance travelled, u = initial velocity, a = acceleration, t = time.

s = 7*3 +(1/2)*32*9 = 165 ft., so the stone is 725 - 165 = 560 ft off the ground.

2. The stone hits the ground when s = 725, so: 725 = 7*t +(1/2)*32*t²

Solving this quadratic we get 

$${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{725}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\
{\mathtt{t}} = {\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = -{\mathtt{6.953\: \!759\: \!395\: \!873\: \!178\: \!7}}\\
{\mathtt{t}} = {\mathtt{6.516\: \!259\: \!395\: \!873\: \!178\: \!7}}\\
\end{array} \right\}$$

The only positive solution is t = 6.516 seconds

3. Use v² = u² + 2as  where v is final velocity.  

v² = 49 + 2*32*725

$${\mathtt{v}} = {\sqrt{{\mathtt{49}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{32}}{\mathtt{\,\times\,}}{\mathtt{725}}}} \Rightarrow {\mathtt{v}} = {\mathtt{215.520\: \!300\: \!667\: \!941\: \!719\: \!5}}$$

v = 215.52 ft/sec

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A litre of water occupies 1000 cm³. The density of ice is about 0.92 times the density of water so it would occupy about 1000/0.92 ≈ 1087 cm³.

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If the result of your calculation with pi is usd in another calculation and that is used in another calculation and that is used in yet another calculation ... etc. rounding errors can build up rapidly, possibly leading to a nonsensical final answer.

Of course, you rarely come across this problem in school level calculations, but in real, industrial, engineering design calculations, for example, this can often be the case.  More digits of pi (and other constants and variables) then need to be carried through the calculations so that rounding errors are kept very small.

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Now Heureka says that AH=HB  

This certainly looks right but at present I can't see how to prove it.

Given that AH = r/2, AD = r and DB = 2r  where r is the radius, then HB = DB - AH - AD = 2r - r/2 - r  = r/2 = AH

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My solution below is basically the same as Chris's, but laid out slightly differently:

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Let x = 0.63333....

Multiply x by 100:  100x = 63.3333....

Multiply x by 10:      10x = 6.3333....

Subtract 10x from 100x:  90x = 63 - 6 or 90x = 57 

Divide both sides by 90:   x = 57/90

Simplify by dividing top and bottom by3:  x = 19/30

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If it's a right-angled triangle the sum of the squares of the two shorter sides will equal the square of the longest side (Pythagoras).

$${{\mathtt{7}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{9}}}^{{\mathtt{2}}} = {\mathtt{130}}$$

$${{\mathtt{15}}}^{{\mathtt{2}}} = {\mathtt{225}}$$

So, clearly not a right-angled triangle

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