A.1.3 Projectile motion Notes

Horizontal and Vertical Components: Separating and Analyzing Independent Motion Components

1. Projectile motion involves an object moving in two dimensions.
2. Yet, until now we have only studied motion in one dimension, thus, it is reasonable to reduce the 2D problem to 1D one.
3. To understand this motion, we break it down into two independent components:
   1. Horizontal motion (constant velocity)
   2. Vertical motion (constant acceleration due to gravity)

Note

• To separately consider horizontal and vertical motion, we, first of all, define motion axis.
• Let O$x$ axis be directed along the ground (horizontally)
• Let O$y$ axis perpendicularly to the ground (vertically upwards).
• If we have a projectile with its launching velocity $\vec{v}$ directed at an angle $\theta$ with respect to O$x$ axis then the project velocity along the axis:
  • $v_x = |\vec{v}|\cos\theta$
  • $v_y = |\vec{v}|\sin\theta$

Horizontal Motion

1. In the horizontal direction, a projectile moves with constant velocity because there is no acceleration (assuming air resistance is negligible), as there is no force acting horizontally.
2. The horizontal displacement ($x$) is given by: $$x = v_{x} \cdot t$$

where $v_{x}$ is the horizontal component of the initial velocity and $t$ is the time of flight.

Vertical Motion

1. In the vertical direction, the projectile is influenced by gravity, which causes a constant downward acceleration of $g = 9.81 \, \mathrm{m/s^2}$.
2. The vertical motion is described by the equations of uniformly accelerated motion:
   1. Vertical velocity at time $t$: $$v_{y} = u_{y} - gt$$ where $u_y = v_y(t=0)$
   2. Vertical displacement at time $t$: $$y = u_{y}t - \frac{1}{2}gt^2$$

Parabolic Trajectories: Describing the Path of a Projectile Under Gravity

1. We will now derive the trajectory of the projectile.
2. Assume a particle is projected with an initial speed $u$ at an angle $\theta$ above the horizontal.
3. We take the origin as the launch point, with:
   1. $u$ = initial speed
   2. $\theta$ = angle of projection (above the horizontal)
   3. $g$ = acceleration due to gravity
   4. $(x, y)$ = position of the object at time $t$
4. The initial velocity components are:
   1. $u_x = u \cos \theta$
   2. $u_y = u \sin \theta$ $$\begin{cases}
      x(t) = u \cos \theta t \\
      y(t) = u \sin \theta t - \dfrac{1}{2} g t^2
      \end{cases}$$
5. To find the equation of the projectile's path \( y(x) \), we eliminate time $t$ from the system.
6. From the horizontal equation: $$t = \frac{x}{u \cos \theta}$$
7. Substitute this into the vertical equation: $$y = u \sin \theta \left( \frac{x}{u \cos \theta} \right) - \frac{1}{2} g \left( \frac{x}{u \cos \theta} \right)^2$$
8. Simplify each term: