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
(1)
$$v_x = u_x$$
(2)
\begin{align} \Delta x = u_x t \end{align}
(3)
$$v_y = u_y + a_y t$$
(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.
• 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.
• 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.
• 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 $\theta = 45^{\circ}$
• 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