### Overarch 2

Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this?

### Stonehenge

Explain why, when moving heavy objects on rollers, the object moves twice as fast as the rollers. Try a similar experiment yourself.

### Maximum Flow

Given the graph of a supply network and the maximum capacity for flow in each section find the maximum flow across the network.

# Dam Busters 1

##### Age 16 to 18 Challenge Level:

Well done to Nathan from South Island school for spotting that a factor of g was missing from the equation with the help of dimensional analysis, and then solved the problem- thought through like a true mathematician!

Sam from JPS found that the bomb would strike at 77 cm from the base. Danielle from Sutton showed that the bomb can strike at most 77cm above the ground. She gave a good expanation, and noted that

I was at first confused by the inequalities but then realised that the plane must be going as fast as it can as high as it can and drop the bomb as near to the dam as possible to get the maximum height.

To put the numbers in the equation I must be using the right units of metres and seconds

$1\mathrm{km} = 1000\mathrm{m}$

$800 \mathrm{km/h} = 800 \times 1000 / 3600\mathrm{m/s}$ because there are $3600$ seconds in $1$ hour and $1000\mathrm{m}$ in $1$ metre.

This is $222\mathrm{m/s}.$

Putting these numbers in shows that the bomb drops $99 \mathrm{m}$ and $23\mathrm{cm}$ by the time it reaches the dam, so it hits at $77\mathrm{cm}$ above the ground.

Ryan from Cornwall also found that the bomb would strike 77cm from the base of the dam, but also realised that this was an ideal situation and that air resistance would change the results.

In reality the bomb would have to pass through the air which would cause friction. This would slow the bomb down. This would slow it down both horizontally and vertically but I think that the biggest effect would be horizontally. Because $77\mathrm{cm}$ is so close to the ground I think that a real bomb would not reach the dam. Also a real bomb would be very big and might hit the ground near to the base of the dam, although its centre would be on course to hit the dam.

Doug provided this full solution:

Neglecting air resistance, which is reasonable since the bomb is small and heavy, and assuming $g$ is constant (almost always reasonable), we can apply the constant acceleration formulae. We know that the horizontal velocity will be constant at $V$ after the bomb is released, so we know the time $t$ to reach the dam after travelling distance $D$ is $t = \frac{D}{V}$. We can therefore consider the vertical motion in this time period.

Using $s = ut + \frac{1}{2}at^2$ (downwards direction positive) with initial velocity $u = 0$, final velocity $v=V$ and acceleration $a = g$ gives the distance dropped $s$ as $s= \frac{1}{2}g \left( \frac{D}{V}\right)^2$.

From the diagram, the minimum dropped distance $s_{min} = H-T$, and the maximum $s_{max} = H-B$. Subsituting the expression for $s$ above to form two inequalities, we can see that $H -\frac{1}{2} g \left( \frac{D}{V} \right)^2 < T$, and $H - \frac{1}{2}g \left( \frac{D}{V} \right)^2 > B$ must both be satisfied.

An additional assumption is that the dam is negligibly thin (i.e. we are not going to hit the top of it!)

For the second part we simply insert the values, and the middle term in the inequality evaluates to $100.775\mathrm{\ m}$, which does (just) satisfy the inequality.

To find the highest point, we see this occurs when the 'plane is highest and nearest (i.e. at $H_{max}$ and $D_{min}$ as before), and again this evaluates to $100.775\mathrm{\ m}$, i.e. just $77.5\mathrm{\ cm}$ above the base of the dam.

Since this margin of error is so slim, we now realise that air resistance and direction and speed of wind will be a consideration.

Extension 1: Show that the change in g is negligible

$$g_1 = \frac{GM_E}{\left(R+\Delta H \right)^2} \textrm{ and } g_2 = \frac{GM_E}{R^2}$$

where $G$ is the Universal Gravitational Constant, $M_E$ the mass of the Earth, $R$ the separation from the centre of Earth at ground level and $\Delta H$ the change in height of the bomb.

The fractional change in $g$ is therefore:

$$\frac{g_2 - g_1}{g_1} = \frac{g_2}{g_1} - 1 = \frac{\left( R + \Delta H \right) ^2}{R^2} - 1\;,$$

$R \approx 6000 \mathrm{\ km}$ (radius of the Earth) and $\Delta H \approx 200 \mathrm{\ m}$ so this evaluates to

$$\frac{(6000 + 0.2)^2}{6000^2} - 1 \approx 0.007\%\;.$$

Extension 2: Show that objects under gravity fall in parabolic curves

A parabola has the general form $y \propto x^2$.

If we let $y$ be the vertical displacement of the bomb (downwards positive) and $x$ the horizontal then the our results above show that $y = \frac{1}{2}at^2$ and $t = \frac{x}{v}$, so $y = \frac{g}{2 v^2}x^2$, which is a parabola.