Non-linear Motion

Table of Contents

• Projectile Motion
  • Analyzing Projectile Motion
  • Examples
• Circular Motion

Projectile Motion

• Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.

• Note that motions along perpendicular axes are independent and thus can be analyzed separately. The key to analyzing a two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical axis.

Analyzing Projectile Motion

• Step 1 Resolve the motion into horizontal and vertical components along the $x$- and $y$-axes. The magnitude of the components of displacement $s$ along these axes are $x$ and $y$. The magnitudes of the components of the velocity $v$ are $v_x=v\cos \theta$ and $v_y=v\sin \theta$, where $v$ is the magnitude of the velocity and $\theta$ is the angle with the positive $x$-axis. Note that initial values are denoted with a subscript $0$. For example, the magnitude of the initial velocity is $v_0$ and its $x$- and $y$- components are:

  $\begin{matrix} v_{0x} = v_0\cos\theta \\ v_{0y} = v_0\sin\theta \end{matrix}$

  
  

  

• Step 2 Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

  Horizontal Motion

  $$\boxed{x = x_0 + v_xt}$$

  (1)

  $$\boxed{v_x = v_{0x}= \text{constant} }$$

  (2)

  Vertical Motion (assume $+$ upward and $g = 9.81~\text{m/s}^2$)

  $$\boxed{a_y = -g = -9.81~\text{m/s}^2}$$

  (3)

  $$\boxed{y = y_0 + v_{0y}t - \frac{1}{2}gt^2 }$$

  (4)

  $$\boxed{v_y = v_{0y} - gt}$$

  (5)

  $$\boxed{v_y^2 = v_{0y}^2 - 2g(y-y_0)}$$

  (6)

• Step 3 Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time $t$.

• Step 4 Recombine quantities in the $x$- and $y$-directions to find the total displacement $\bm{s}$ and velocity $\bm{v}$. Solve for the magnitude and direction of $\bm{s}$ and $\bm{v}$ using:

  $$\boxed{s = \sqrt{x^2 + y^2}}$$

  (7)

  $$\boxed{\theta = \tan^{-1}\left(\frac{y}{x}\right) }$$

  (8)

  $$\boxed{v = \sqrt{v_x^2 + v_y^2}}$$

  (9)

  $$\boxed{\theta_v = \tan^{-1}\left(\frac{v_y}{v_x}\right) }$$

  (10)

Examples

Circular Motion