$\vec{r_{1}}=x_{1}\,\hat{i}+y_{1}\,\hat{j}+z_{1}\,\hat{k}$

$\vec{r_{2}}=x_{2}\,\hat{i}+y_{2}\,\hat{j}+z_{2}\,\hat{k}$

$\Delta\vec{r}=\vec{r_{2}}-\vec{r_{1}}=(x_{2}-x_{1})\,\hat{i}+(y_{2}-y_{1})\,\hat{j}+(z_{2}-z_{1})\,\hat{k}$

Average velocity: $\vec{v_{ave}}=\frac{\Delta\vec{r}}{\Delta t}$

Instantaneous velocity: $\vec{v}=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{i}+\frac{dy}{dt}\,\hat{j}+\frac{dz}{dt}\,\hat{k}=v_{x}\,\hat{i}+v_{y}\,\hat{j}+v_{z}\,\hat{k}$

Average acceleration: $\vec{a_{ave}}=\frac{\Delta\vec{v}}{\Delta t}$

Instantaneous acceleration: $\vec{a}=\frac{d\vec{v}}{dt}=\frac{dv_x}{dt}\,\hat{i}+\frac{dv_y}{dt}\,\hat{j}+\frac{dv_z}{dt}\,\hat{k}=\frac{d^{2}x}{dt^2}\,\hat{i}+\frac{d^{2}y}{dt^2}\,\hat{j}+\frac{d^{2}z}{dt^2}\,\hat{k}$

At the moment the hunter pulls the trigger the monkey can make a decision, should he hang on or drop from the tree in the hope the bullet will miss him?

In case my monkey shooting fails Physclips has a video you can step through frame by frame and a pretty detailed explanation here

Height of bullet with time:

$y=x\tan\theta-\frac{1}{2}gt^{2}$

Hunter's expectation:

$y=x\tan\theta$

At monkey's $x$ position, $x_{m}$ the bullet will be $-\frac{1}{2}gt^{2}$ below where monkey started…exactly the same distance the monkey will be below his starting point if he lets go.

This is a simple demonstration of the fact that gravity accelerates all objects equally independently of their horizontal velocity.

If $g=9.8ms^{-2}$

$\Large a_{x}=0$ | $\Large a_{y}=-g$ |

$\Large v_{x}=v_{x0} $ | $\Large v_{y}=v_{y0}-gt$ |

$\Large x=x_0+v_{x0} t$ | $\Large y=y_{0}+v_{y0} t -\frac{1}{2}gt^{2}$ |

$\Large a_{x}=0$ | $\Large a_{y}=-g$ |

$\Large v_{x}=v_{0}\cos\theta$ | $\Large v_{y}=v_{0}\sin\theta-gt$ |

$\Large x=x_0+v_{0}\cos\theta t$ | $\Large y=y_{0}+v_{0}\sin\theta t -\frac{1}{2}gt^{2}$ |

Taking $x_{0}$,$y_{0}$ as $(0,0)$

$t=\frac{x}{v_{0}\cos\theta}$

$y=\frac{\sin\theta}{\cos\theta}x-\frac{g}{(2v_{0}^{2}\cos^2\theta)} x^{2}$

Horizontal Range → Solve for $y=0$

$0=x(\frac{\sin\theta}{\cos\theta}-\frac{g}{(2v_{0}^{2}\cos^2\theta)}x)$

$x=0$ and $x=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$ → $R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$

0.04/.0095=4.2m/s

Note that in terms of the equations shown previously $\theta=(180^o-angle)$ as the motion is from right to left.

$R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}=\frac{v_{0}^{2}}{g}\sin 2\theta$

$\frac{dR}{d\theta}=\frac{2v_{0}^2}{g}(\frac{d(\sin\theta)}{d\theta}\cos\theta)+\sin\theta\frac{d(\cos\theta)}{d\theta})$

$=\frac{2v_{0}^2}{g}(\cos^2\theta-\sin^2\theta)=\frac{2v_{0}^2}{g}\cos2\theta$

or

$\frac{dR}{d\theta}=\frac{v_{0}^2}{g}\frac{d(\sin{2\theta})}{d\theta}\frac{d(2\theta)}{d\theta}=\frac{2v_{0}^2}{g}\cos2\theta$

$\frac{dR}{d\theta}=0$ at $\theta=45^o,135^o$

$45^o$ and $135^o$ correspond to maximum range.

From our equations

$\Large a_{x}=0$ | $\Large a_{y}=-g$ |

$\Large v_{x}=v_{x0} $ | $\Large v_{y}=v_{y0}-gt$ |

$\Large x=x_0+v_{x0} t$ | $\Large y=y_{0}+v_{y0} t -\frac{1}{2}gt^{2}$ |

we can derive that horizontal distance traveled is

$x= v_{x0}\Large \sqrt{\frac{2h}{g}}$

If we consider two points on a turning disc at distance $r_{1}$ and $r_{2}$ from the center of the disc we can make a number of observations.

First the distance traveled by the points is quite different, points that are far from the center go through a greater distance.

However, the angular displacement is the same for both points, suggesting that angular displacement is a good variable to describe the motion of the whole rotating object.

We would like to be able to have a straight forward relationship between the actual distance traveled and the angular displacement, which we can have if we express our angular displacement in radians.

As the relationship between the arc length on a circle and the angle which it subtends in radians is

$l=r\theta$

a change in the angular displacement of $\Delta\theta$ results in a change in the displacement

$\Delta l=r\Delta\theta$

For change in angular displacement $\Delta \theta$ in a time interval $\Delta t$ we can define an average angular velocity

$\bar{\omega}=\frac{\Delta \theta}{\Delta t}$

As we are now accustomed, we can also define an instantaneous velocity

$\omega=\lim_{\Delta t\to 0}\frac{\Delta \theta}{\Delta t}=\frac{d\theta}{dt}$

The units of angular velocity can be expressed as either $\mathrm{rad\,s^{-1}}$ or $\mathrm{s^{-1}}$

The tangential and angular velocity are related through $v=\omega r$

Frequency (revolutions per second) $f$ [ $s^{-1}$ or $\mathrm{Hz}$]

Period (time for one revolution) $T=\frac{1}{f}$ [$\mathrm{s}$]

Velocity $v=\frac{2\pi r}{T}=2\pi r f$ [$ms^{-1}$]

Angular Velocity $\omega=\frac{2\pi}{T}=2\pi f$ [$s^{-1}$]

The relationship between frequency and period can be explored here.

The magnitude of the velocity does not change, but it's direction does. The vector for the change of the velocity always points towards the center of the circle. There is thus an acceleration, named the **centripetal acceleration** pointing to the center of the circle. For small changes in the position.

$\Delta v=2v\sin(\frac{\Delta\theta}{2})\approx v \Delta\theta$

$\Delta\theta=\frac{\Delta l}{r}$

$v\Delta\theta=v\frac{\Delta l}{r}$

$a_{R}=\lim_{\Delta t\to 0}\frac{\Delta v}{\Delta t} = \lim_{\Delta t\to 0} \frac{v}{r}\frac{\Delta l}{\Delta t}=\frac{v^2}{r}=\omega^{2}r$ [$ms^{-2}$]

Dealing with relative velocity is a particularly important application of addition and subtraction of vectors.

We can adopt a notation which can be helpful.

$\vec{v}_{AB}=\vec{v}_{A}-\vec{v}_{B}$: velocity of A relative to B

If we want to know the velocity of A relative to C if we know the velocity of A relative to B and the velocity of B relative to C

$\vec{v}_{AC}=\vec{v}_{AB}+\vec{v}_{BC}$

Whereas if we want to find the velocity of B relative to C

$\vec{v}_{BC}=\vec{v}_{AC}-\vec{v}_{AB}$