# physics

## Tow-dimentional projectile motion

### $$\vec{a} + \vec{b} = \vec{c}$$ --------- $$\vec{x}_{h}:x \quad horisontal$$ $$\vec{x}_{v}:x \quad vertical$$ $$\vec{x} = \vec{x}_{h} + \vec{x}_{v}$$

$$displacement:\quad \| \vec{a} \|=5$$ $$angle \quad \theta = 36.8699°$$ $$\| \vec{a}_{x} \| = \| \vec{a} \| \centerdot cos\theta = 4$$ $$\| \vec{a}_{y} \| = \| \vec{a} \| \centerdot sin\theta = 3$$

### Projectile an angle

Lauanch a rocket $$\quad velocity:\vec{v}=10m/s$$ $$\quad direction:\theta= 30° \quad upward \quad from \quad the \quad ground$$
How far does it travel?
$$\| \vec{v}_{y} \| :$$ $$\quad sin30° = \frac{\|\vec{v}_{y}\|}{\|\vec{v}\|} \rightarrow \|\vec{v}_{y}\|=\|\vec{v}\|sin30°=10m/s \centerdot \frac{1}{2}=5m/s$$ $$\| \vec{v}_{x} \|:$$ $$\quad cos30° = \frac{\|\vec{v}_{x}\|}{\|\vec{v}\|} \rightarrow \|\vec{v}_{x}\|=\|\vec{v}\|cos30°=10m/s \centerdot \frac{\sqrt{3}}{2}=5\sqrt{3}m/s$$ $$change \quad in \quad velocity:\Delta \vec{v}_{g} =\vec{v}_{f}-\vec{v}_{i}=\vec{a}_{g} \centerdot \Delta t$$ $$\quad initial \quad velocity = -final \quad velocity$$ $$\quad \Delta \vec{v}_{g}=\vec{v}_{f}-\vec{v}_{i} =-5m/s-5m/s=-10m/s$$ $$change \quad in \quad time\quad (time \quad in \quad air): \Delta t = \frac{\vec{v}_{g}}{\vec{a}_{g}}$$ $$\quad acceleration:\vec{a}_{g}=-9.8m/s^2$$ $$\quad \Delta t = \frac{-10m/s}{-9.8m/s^2} =1.02s$$ $$displacement(how \quad far \quad a \quad rocket \quad travels): =\vec{v}_{x} \centerdot \Delta t$$ $$\quad \vec{s}_{x}=5\sqrt{3}m/s \centerdot 1.02s = 8.83m$$
A rocket travels 8.83 meters.

### Different way to determine time in air

$$vertical \quad velocity: \vec{v}_{y}=5m/s$$ $$displacement:\vec{s}=\vec{v}_{average} \centerdot \Delta t$$ $$average \quad velocity:\vec{v}_{avg}=\frac{\vec{v}_{i}+\vec{v}_{f}}{2}$$ $$final \quad velocity:\vec{v}_{f}=\vec{v}_{i}+\vec{a} \centerdot \Delta t$$

$$\vec{s}=\frac{\vec{v}_{i}+\vec{v}_{i}+\vec{a} \centerdot \Delta t}{2}\centerdot\Delta t$$

$$\vec{s}=\vec{v}_{i}\centerdot \Delta t + \frac{\vec{a}}{2}\centerdot (\Delta t)^2$$

$$\vec{s}=5m/s \centerdot \Delta t + -4.9m/s^2 \centerdot (\Delta t)^2$$ $$vertical \quad displacement: 0m \nearrow peak \searrow 0m$$ $$0=5m/s \centerdot \Delta t + -4.9m/s^2 \centerdot (\Delta t)^2$$ $$0=\Delta t(5m/s-4.9m/s^2 \centerdot \Delta t)$$

This equation equals to 0 when $$\Delta t=0$$ and $$5m/s-4.9m/s^2 \centerdot \Delta t=0$$ $$\Delta t=\frac{-5m/s}{-4.9m/s^2}$$

$$\Delta t=1.02s$$

### Launching and landing on different elevations

Launch a rocket $$\quad velocity:90m/s$$ $$\quad direction:53°$$ How far does a rocket travel? $$\|\vec{v}_{y}\|=90sin53°$$ $$\|\vec{v}_{x}\|=90cos53°$$
vertical displacement (acceleration) $$\vec{s}_{y}=\vec{v}_{i}\centerdot \Delta t + \frac{\vec{a}}{2}\centerdot (\Delta t)^2$$ horisontal displacement $$\vec{s}_{x}=\vec{v}_{i}\centerdot \Delta t$$

$$-16=(90sin53°)\centerdot \Delta t - 4.9(\Delta t)^2 \rightarrow -4.9(\Delta t)^2 + (90sin53°)\centerdot \Delta t + 16 = 0$$

$$ax^2 + bx + c = 0 \rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

$$\Delta t = \frac{-(90sin53°)\pm\sqrt{(90sin53°)^2-4\times (-4.9)\times 16}}{2\times(-4.9)}$$ time is positive $$\Delta t = \frac{-(71.8772)-\sqrt{5166.3319+313.6}}{-9.8}=14.89s$$ horisontal displacement is $$\vec{s}_{x}=90cos53°\times 14.89=806.5m$$

### Total displacement for projectile

Launch a rocket $$\quad velocity:30m/s$$ $$\quad direction:80°$$ How far does a rocket travel? $$\|\vec{v}_{y}\|=30sin80°=29.54m/s$$ $$\|\vec{v}_{x}\|=30cos80°=5.21m/s$$ Vertical displacement $$\vec{s}_{y}=\vec{v}_{y}\centerdot \Delta t + \frac{a}{2}(\Delta t)^2$$ $$\vec{s}_{y}=10m$$ $$10m=30sin80°m/s\centerdot \Delta t -4.9m/s^2\centerdot (\Delta t)^2$$ $$-4.9m/s2\centerdot (\Delta t)^2 + 29.54m/s\centerdot \Delta t -10 = 0$$ $$\Delta t = \frac{-29.54\pm\sqrt{29.54^2-4(-4.9)(-10)}}{-9.8}=5.67s$$ Horizontal displacement $$\vec{s}_{x}=\vec{v}_{x}\centerdot \Delta t$$ $$\quad =30cos80°m/s\centerdot 5.67s=5.21m/s\centerdot 5.67s=29.53m$$
Total displacement $$\vec{s}=\vec{s}_{x}+\vec{s}_{y}$$ $$\|\vec{s}\|=\sqrt{(\vec{s}_{x})^2+(\vec{s}_{y})^2} =\sqrt{10^2+29.53^2}=31.18m$$

### Total final velocity for projectile

Vertical chage in velocity $$\vec{v}_{y}=\vec{a}_{y}\centerdot \Delta t$$ $$\vec{v}_{f}-\vec{v}_{i}=\vec{a}_{y}\centerdot \Delta t$$ $$\vec{v}_{f}-29.54m/s=-9.8m/s^2\centerdot 5.67s$$ $$\vec{v}_{f}=29.54m/s-9.8m/s^2\centerdot 5.67s \quad = -26.03m/s\quad(downward)$$ Total final velocity $$\|\vec{v}\|=\sqrt{(\vec{v}_{x})^2+(\vec{v}_{y})^2} =\sqrt{5.21+26.03^2}=26.55m/s$$ $$tan\theta = \frac{26.03}{5.21}$$ $$\theta = tan^{-1}(\frac{26.03}{5.21})$$ $$\quad = -78.7 \quad (below \quad horisontal)$$

### Projectile on an incline

an incline $$\quad degrees:30°$$ projectile $$\quad direction:15°$$ $$\quad velocity:10m/s$$  $$\|\vec{v}_{y}\|=10sin(15°+30°)=\frac{\sqrt{2}}{2}m/s$$ $$\|\vec{v}_{x}\|=10cos(15°+30°)=\frac{\sqrt{2}}{2}m/s$$ vertical displacement $$\vec{s}_{y}=\vec{v}_{y}\centerdot \Delta t + \frac{a}{2}\centerdot (\Delta t)^2$$ $$\quad = 10sin45°\centerdot \Delta t + \frac{-9.8}{2}\centerdot (\Delta t)^2$$ $$\quad = 5\sqrt{2}\centerdot \Delta t - 4.9\centerdot (\Delta t)^2$$ horizontal displacement $$\vec{s}_{x}=\vec{v}_{x}\centerdot \Delta t$$ $$\quad = 10cos45°\centerdot \Delta t$$ $$\quad = 5\sqrt{2}\centerdot \Delta t \rightarrow \Delta t=\frac{\vec{s}_{x}}{5\sqrt{2}}$$ tangent $$tan30°=\frac{\|\vec{s}_{y}\|}{\|\vec{s}_{x}\|} =\frac{sin30°}{cos30°}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} =\frac{1}{\sqrt{3}}$$ $$\frac{\|\vec{s}_{y}\|}{\|\vec{s}_{x}\|}=\frac{1}{\sqrt{3}} \rightarrow \vec{s}_{x}=\sqrt{3}\vec{s}_{y}$$ $$\Delta t=\frac{\vec{s}_{x}}{5\sqrt{2}} =\frac{\sqrt{3}\vec{s}_{y}}{5\sqrt{2}}$$ vertical displacement $$\vec{s}_{y}=5\sqrt{2}\centerdot \Delta t - 4.9\centerdot (\Delta t)^2$$ $$\quad=5\sqrt{2}\centerdot \frac{\sqrt{3}\vec{s}_{y}}{5\sqrt{2}} - 4.9\centerdot (\frac{\sqrt{3}\vec{s}_{y}}{5\sqrt{2}})^2$$ $$\quad=\sqrt{3}\vec{s}_{y}-\frac{4.9\times 3}{50}\vec{s}_{y}^2$$ $$\downarrow$$ $$0=\sqrt{3}\vec{s}_{y}-\vec{s}_{y}-\frac{4.9\times 3}{50}\vec{s}_{y}^2$$ $$0=(\sqrt{3}-1)\vec{s}_{y}-\frac{4.9\times 3}{50}\vec{s}_{y}^2$$ $$0=((\sqrt{3}-1)-\frac{4.9\times 3}{50}\vec{s}_{y})\vec{s}_{y}$$ $$\vec{s}_{y}=0 \quad or$$ $$(\sqrt{3}-1)-\frac{4.9\times 3}{50}\vec{s}_{y}=0$$ $$\vec{s}_{y}=(\sqrt{3}-1)\cdot \frac{50}{4.9\times 3}=2.49m$$ $$\vec{s}_{x}=\sqrt{3}\cdot \vec{s}_{y}=\sqrt{3}\cdot2.49=4.31m$$

### Unit vectors and engineering notation

$$\vec{s}_{y}=10\cdot sin(30°)=10\cdot\frac{1}{2}=5$$ $$\vec{s}_{x}=10\cdot cos(30°)=10\cdot\frac{\sqrt{3}}{2}=5\sqrt{3}$$ define unit vectors
vertical unit vector $$\quad \hat{j}=\vec{1} \rightarrow \vec{s}_{y}=5\hat{j}$$ horizontal unit vector $$\quad \hat{i}=\vec{1} \rightarrow \vec{s}_{x}=5\sqrt{3}\hat{i}$$

### Unit vector notation

horisontal unit vector $$\hat{i}=1$$ vertical unit vector $$\hat{j}=1$$ unit vector notation $$\vec{a}= 10\hat{i} + 3\hat{j}$$ $$\vec{b}= 2\hat{i} + 4\hat{j}$$ $$\vec{a}+\vec{b}= (10+2)\hat{i} + (3+4)\hat{j}$$ $$\vec{a}+\vec{b}= 12\hat{i} + 7\hat{j}$$

### Unit vector notation(part 2)

$$\vec{a}=-3\hat{i} + 2\hat{j}$$ $$\vec{b}=2\hat{i} + 4\hat{j}$$ $$\vec{a}+\vec{b}=(-3+2)\hat{i} + (2+4)\hat{j}=-1\hat{i}+6\hat{j}$$

### Projectile motion with ordered set notation

unit vector notation $$\vec{a}=-3\hat{i} + 2\hat{j}$$ ordered set notation $$\vec{a}=<-3,2>$$

I hit a ball $$\quad 4 \quad feet \quad high \quad above \quad the \quad ground$$ $$\quad velocity \quad is \quad |\vec{v}|=120ft/s, \quad direction \quad 30° \quad to \quad the \quad horisontal$$

There is a fence $$\quad350 \quad feet \quad away, \quad and \quad 30 \quad feet \quad tall$$

acceleration $$\quad \vec{a}=-32 ft/s^2$$ wind gust $$velocity=5ft/s \quad to \quad the \quad right$$

Is the ball going to clear the fence?

initial position of a ball $$\vec{p}_{i}=<0,4>$$

$$\vec{p}(t)=\vec{p}_{i}+\vec{v}_{i}\cdot t + \frac{\vec{a}}{2}\cdot t^2$$ initial velocity $$\vec{v}_{x}=120 \cdot cos30° + 5(gust)=120 \cdot \frac{\sqrt{3}}{2}+5=109\qquad \vec{v}_{y}=120 \cdot sin30°=120 \cdot \frac{1}{2}=60$$ $$\vec{v}_{i}=<109,60>$$ $$\vec{a}=<0,-32>$$

$$\vec{p}(t)=<0,4>+<109,60>t+\frac{t^2}{2}<0,-32>$$ $$\vec{p}(t)=<0,4>+<109t,60t>t+<0,-16t^2>$$ $$\vec{p}(t)=<109t,4+60t-16t^2>$$

time $$350ft = \vec{v}_{x}ft/s\cdot t \rightarrow t= \frac{350ft}{109ft/s}=3.2s$$ $$\vec{p}(3.2)=<109\times3.2,4+60\times3.2+16\times(3.2)^2>$$ $$\quad=<350,32>$$