Projectile motion Knowledge-Based Study

Study notes

1. What is Projectile Motion?

   Projectile motion occurs when an object is launched into the air and moves under the influence of gravity alone, assuming air resistance is negligible. The path traced by a projectile is called its trajectory, which forms a parabola.

   Independence of Horizontal and Vertical Components

   The foundational principle of two-dimensional kinematics is that perpendicular motion components act independently of one another:

   • Horizontal Motion ($x$-axis): There is no horizontal force acting on the projectile (neglecting air resistance). Therefore, horizontal acceleration is zero ($a_x = 0$), and the horizontal velocity remains constant throughout the entire flight ($v_x = v_{0x}$).
   • Vertical Motion ($y$-axis): Gravity acts continuously downward. Thus, the vertical acceleration is constant and directed downward ($a_y = -g$, where $g \approx 9.8\text{ m/s}^2$). The vertical motion is identical to standard free fall.

   Because these two motions occur simultaneously, the single quantity linking both directions is time ($t$).

   Analyzing Horizontal Launches

   When an object is launched horizontally from an initial height $h$ with an initial speed $v_0$:

   • Initial horizontal velocity: $v_{0x} = v_0$
   • Initial vertical velocity: $v_{0y} = 0$

   Equations of Motion

   Taking the launch point as the origin and defining upward as positive:

   1. Vertical Analysis:

   The vertical displacement is $\Delta y = -h$. Using the kinematic relation:

   $$\Delta y = v_{0y}t - \frac{1}{2}gt^2 \implies -h = -\frac{1}{2}gt^2$$

   Solving for the time of flight ($t$):

   $$t = \sqrt{\frac{2h}{g}}$$

   Notice that the time of flight depends exclusively on the vertical height and the acceleration due to gravity, completely independent of the horizontal launch speed $v_0$.

   2. Horizontal Analysis:

   The horizontal distance traveled, known as the range ($R$), is determined by constant velocity:

   $$R = v_{0x}t = v_0 \sqrt{\frac{2h}{g}}$$

   Instantaneous Velocity

   At any time $t$, the vertical velocity is $v_y = -gt$. The magnitude of the total velocity is given by:

   $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + (-gt)^2}$$

   Decomposing Initial Velocity

   When a projectile is launched with an initial velocity $v_0$ at an angle $\theta$ above the horizontal, resolve the velocity vector into rectangular components:

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

   Key Characteristics of Symmetrical Trajectories

   For a projectile launched from and returning to the same vertical level on flat ground:

   • Time to Peak ($t_{\text{peak}}$): At the highest point of the trajectory, the vertical velocity is momentarily zero ($v_y = 0$). Using $v_y = v_{0y} - gt$:
     $$t_{\text{peak}} = \frac{v_0 \sin\theta}{g}$$
   • Total Flight Time ($T$): Because the upward and downward phases are symmetric under constant gravity:
     $$T = 2 t_{\text{peak}} = \frac{2v_0 \sin\theta}{g}$$
   • Maximum Height ($H$): Using the kinematic formula $v_y^2 = v_{0y}^2 - 2gH$:
     $$H = \frac{(v_0 \sin\theta)^2}{2g}$$
   • Horizontal Range ($R$): The total horizontal displacement is $R = v_{0x} T$:
     $$R = (v_0 \cos\theta)\left(\frac{2v_0 \sin\theta}{g}\right) = \frac{v_0^2 \sin(2\theta)}{g}$$

   Critical Conceptual Takeaways

   • Velocity at Maximum Height is Not Zero: At the apex of trajectory, only the vertical component of velocity is zero ($v_y = 0$). The horizontal component remains unchanged ($v_x = v_0 \cos\theta$), so the total velocity at the peak is strictly non-zero (unless the launch angle was exactly $90^\circ$).
   • Acceleration is Constant Everywhere: At all points along the path—including the top of the arc—the acceleration vector is constant, directed downward, and has magnitude $g$. An object momentarily at zero vertical velocity at the top still experiences full gravitational acceleration.
   • Complementary Launch Angles: For a given launch speed over level ground, angles that sum to $90^\circ$ (such as $30^\circ$ and $60^\circ$) yield identical horizontal ranges, because $\sin(2\theta) = \sin(2(90^\circ - \theta))$. However, the larger angle attains a greater maximum height and spends more time in the air.
   • Maximum Range Condition: On level ground without air resistance, $\sin(2\theta)$ reaches its maximum value of 1 when $2\theta = 90^\circ$, meaning $\theta = 45^\circ$ yields the maximum possible range.