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:
   1. First term: $$u \sin \theta \frac{x}{u \cos \theta} = x \tan \theta$$
   2. Second term: $$\frac{1}{2} g \frac{x^2}{u^2 \cos^2 \theta}
      = \frac{g x^2}{2 u^2 \cos^2 \theta}
      $$
9. Putting it all together: $$y(x) = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$$

Key Features of a Parabolic Trajectory

1. Symmetry: The path is symmetrical about the highest point (the apex).
2. Maximum Height: The vertical velocity is zero at the apex.
3. Range: The total horizontal distance traveled by the projectile.
4. Time of Flight: The total time the projectile is in the air.

Optimal Launch Angles: Understanding the Relationship Between Launch Angle and Range

Derivation of range

1. We have previously obtained the expression for the trajectory of a projectile.
2. Now, one may wonder, how to find its range (how far does it fly), and how to choose a launching angle to maximize the range.
3. We have previously established that:$$y(x) = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$$
4. To find the range, we determine the horizontal distance at which the projectile hits the ground, i.e., when \( y(x) = 0 \): $$x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} = 0$$ $$x\left(\tan \theta - \frac{g x}{2 u^2 \cos^2 \theta} \right) = 0 $$

Now, the first solution is obvious: $x = 0$. But the point $(0, 0)$ is the origin. This solution tells us, that we were on the ground level during the launch of projectile - not very useful.

1. Instead we proceed with finding the second solution: $$\tan \theta - \frac{g x}{2 u^2 \cos^2 \theta} = 0$$
2. Solving for \( x \), we find the range \( R \): $$\frac{g x}{2 u^2 \cos^2 \theta} = \tan \theta$$ $$x = \frac{2 u^2 \cos^2 \theta \tan \theta}{g}$$
3. Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), so: $$x = \frac{2 u^2 \cos^2 \theta \cdot \frac{\sin \theta}{\cos \theta}}{g} = \frac{2 u^2 \cos \theta \sin \theta}{g}$$
4. Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we obtain: $$R = \frac{u^2 \sin 2\theta}{g}$$

We have just obtained that the range ($R$) of a projectile launched with an initial velocity $u$ at an angle $\theta$ is given by: $$R = \frac{u^2 \sin 2\theta}{g}$$

Symmetrical Angles

Two launch angles that add up to $90^\circ$ (e.g., $30^\circ$ and $60^\circ$) will produce the same range, but with different trajectories.

1. Lower angles result in flatter trajectories with shorter flight times.
2. Higher angles result in higher trajectories with longer flight times.