### High Jumping

How high can a high jumper jump? How can a high jumper jump higher without jumping higher? Read on...

### Dam Busters 1

At what positions and speeds can the bomb be dropped to destroy the dam?

### Dam Busters 2

Can you work out which of the equations models a bouncing bomb? Will you be able to hit the target?

# Bike

##### Stage: 5 Challenge Level:

As suggested in the hint it is useful to choose the coordinate system with the origin at the wheel touching the ground point. The wheel is moving horizontally with the speed $v = 18\mathrm{kmh}^{-1} = 5\mathrm{ms}^{-1}$ and the diameter of the wheel $d = 23 \mathrm{inch} = 0.584\mathrm{m}$. Note $r = d/2$. The wheel is not slipping on the ground, thus the linear speed of the drops on the wheel is $v = 5\mathrm{ms}^{-1}$. Suppose that the drop detaches when it is at an angle $\alpha$ with vertical.

In x direction:  $x(t) = -r\sin(\alpha) + vt + vt\cos(\alpha)\;.$

In y direction:  $y(t) = r + r\cos(\alpha) + vt\sin(\alpha) - \frac{gt^2}{2}\;.$

We neglect any air resistance then the drop will be at its highest point when its vertical speed is zero.

$$0 = v\sin(\alpha) - gt_{max}$$

Hence, $t_{max} = \frac{v\sin(\alpha)}{g}$. We substitute this expression to the y direction equation to get that

$$y(\alpha) = r(1 + \cos(\alpha))+\frac{v^2\sin(\alpha)^2}{g}-\frac{v^2\sin^2(\alpha)}{2g} = r(1 + \cos(\alpha)) +\frac{v^2\sin^2(\alpha)}{2g}\;.$$

Method 1. Differentiate y with respect to alpha and then solve $\frac{dy}{d\alpha} = 0$. Remember that $\frac{d\sin(\alpha)}{d\alpha} = \cos(\alpha)$ and  $\frac{d\cos(\alpha)}{d\alpha} = -\sin(\alpha)$. We can use chain rule to differentiate $\sin^2(\alpha)$: $$\frac{d\sin^2(\alpha)}{d\alpha} = \frac{d\sin^2(\alpha)}{d\sin(\alpha)}\frac{d\sin(\alpha)}{d\alpha} = 2\sin(\alpha)\cos(\alpha)\;.$$ Moreover if C is constant $\frac{dC}{d\alpha} = 0$. Thus, $\frac{dy}{d\alpha} = 0 - r\sin(\alpha) + \frac{v^2}{2g}2\sin(\alpha)\cos(\alpha)$. This means that either $\sin(\alpha) = 0$ or $\cos(\alpha) =\frac{gr}{v^2}$ for $\frac{gr}{v^2} < 1$. If $\sin(\alpha) = 0$ then $\cos(\alpha) = 1$ or $-1$, $y_{max} = 2r = 0.584\mathrm{m}$ or $0$. If $\cos(\alpha) =\frac{gr}{v^2}$ for $\frac{gr}{v^2} < 1$ then $y_{max} = r\left(1+\frac{gr}{v^2}\right) + \frac{v^2}{2g}\left(1 -\frac{g^2r^2}{v^4}\right) = r + \frac{v^2}{2g} + \frac{gr^2}{2v^2} = 1.58\mathrm{m}$. So, the maximum height is $1.58\mathrm{m}$.

Method 2. Use identity $\sin^2(\alpha) = 1 - \cos^2(\alpha)$ and note $X = \cos(\alpha)$. Then $y(X) = r + rX + \frac{v^2}{2g} - \frac{v^2}{2g}X^2$ which is a quadratic and the critical point can be found. Try it!

Find the $x$ coordinate of the drops when they are at the highest point.

$$\begin{eqnarray} x_{max} &=& -r\sin(\alpha) + vt(1 + \cos(\alpha)) = -r\sin(\alpha) + v\frac{v\sin(\alpha)}{g}(1 + \cos(\alpha))\\ &=& \sin(\alpha)\left(\frac{v^2}{g}(1 + \cos(\alpha)) - r\right)\\ &=& \sqrt{1 - \frac{g^2r^2}{v^4}}\left(\frac{v^2}{g}\left(1 + \frac{gr}{v^2}\right) -r\right) =\frac{v^2}{g} \sqrt{1 - \frac{g^2r^2}{v^4}} \end{eqnarray}$$

The coordinate of the wheel axle is $x_{axle} = vt = v\frac{v\sin(\alpha)}{g} = \frac{v^2}{g} \sqrt{1 - \frac{g^2r^2}{v^4}}$. Thus, $x_{axle} = x_{max}$. The drops are just above the wheel axle.