Projectile Motion

#### Intro

- A
**projectile**is any object that undergoes**non-powered flight**. (e.g. Ball being kicked) - Upon being fired, the projectile follows a
**trajectory**, moving through the air. - Galileo postulated that this motion of the projectile can be regarded as
**two separate and independent motions**:**Vertical**motion (which we will use the**y**axis to plot)**Horizontal**motion (which we will use the**x**axis to plot)

- To analyse these equations, we take each of the motions
**separately** **NB:**As the motion is**unassisted,**we assume the only force acting on the projectile is**gravity**.

#### Equations of motion

**NB**: For the following:

- $u$ is defined as
**initial velocity** - It is assumed there is
**no force in the horizontal motion**(e.g. no air resistance) - Acceleration in the y axis is
**gravity**

\begin{equation} v_x = u_x \end{equation}

(2)
\begin{align} \Delta x = u_x t \end{align}

(3)
\begin{equation} v_y = u_y + a_y t \end{equation}

(4)
\begin{align} v_y^2 = u_y^2 + 2a\Delta y \end{align}

(5)
\begin{align} \Delta y = u_y t+\frac{1}{2}a_y t^2 \end{align}

#### Step by Step Method

**Step 1:**Define the polarity- Usually, we define
**up**and**right**as**positive.** - This would mean that
**gravity**or $a_y$ is**negative.**

- Usually, we define

**Step 2:**Splitting the components of the initial velocity- As mentioned previously, projectile motion can be split into 2 components on a 2D plane.
- See the picture to see how.
- This means that:
- $u_x = u cos{\theta}$
- $u_y = u sin{\theta}$

**Do this for EVERY projectile motion question. EVERY. SINGLE. ONE.**

**Step 3:**Finding the maximum height- This occurs at $v_y = 0$
- With that and $v_y^2 = u_y^2 + 2a\Delta y$, you can find the max height by
**substitution**. **NB:**For a**flat ground**, this occurs at the**middle**of its trajectory.- That is, the
**vertex**of the parabola, or at**half the time of flight**.

- That is, the

**Step 4:**Finding the time of flight- Use $\Delta y = u_y t+\frac{1}{2}a_y t^2$ where $\Delta y$ is the
**difference in height**of its origin and landing place. - This usually gets you a
**quadratic**. Remember the answer is in the domain: $t > 0$ - For
**even ground**, $\Delta y = 0$ - Also for even ground, time of flight is
**twice**the time taken to reach**max height**.

- Use $\Delta y = u_y t+\frac{1}{2}a_y t^2$ where $\Delta y$ is the
**Step 5:**Finding the range- This is given by: $\Delta x = u_x t$ where
**t is the time of flight**. - For even ground,
**maximum range**is given by $\theta = 45^{\circ}$

- This is given by: $\Delta x = u_x t$ where
**Step 6:**Finding the final velocity- Utilising $v_x = u_x$ and $v_y = u_y + a_y t$, we can find the horizontal and vertical components of the final velocity.
- With that, we can use $tan{\theta} = \frac{v_y}{v_x}$ to find its
**angle** - We can also use
**Pythagoras' Theorem**to find its magnitude.

**NB:** An interesting question that came up was: what would the trajectory of a dropped projectile be like if the projectile's frame of reference was accelerating horizontally.

If this acceleration is constant, it would be a straight line but the horizontal component would be in the opposite direction to the acceleration due to inertia.

page revision: 2, last edited: 03 Oct 2011 07:18