Horizontal launch Knowledge-Based Study

Study notes

1. What is Horizontal Launch?

   A horizontal launch occurs when an object is projected horizontally from a certain elevation above the ground with an initial velocity $v_0$ under the sole influence of gravity (assuming negligible air resistance).

   According to Galileo's principle of the independence of perpendicular motions, a two-dimensional projectile motion can be resolved into two mutually independent one-dimensional motions:

   • Horizontal axis ($x$): Moves with constant velocity because there is no net horizontal force ($a_x = 0$).
   • Vertical axis ($y$): Undergoes uniform acceleration downward due to gravity ($a_y = g$), starting with an initial vertical velocity of zero ($v_{0y} = 0$).

   Analysis of the Vertical Component

   In the vertical direction, the projectile behaves identically to an object released from rest in free fall from a height $h$.

   Taking the downward direction as positive and using the kinematic equation for position:

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

   Since the initial vertical velocity is zero ($v_{0y} = 0$), when the object reaches the ground ($y = h$), the equation simplifies to:

   $$h = \frac{1}{2}gt^2$$

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

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

   Key Insight: The time required for an object launched horizontally to reach the ground depends exclusively on the launch height $h$ and the gravitational acceleration $g$. It is completely independent of the initial horizontal velocity $v_0$.

   Analysis of the Horizontal Component

   Because no horizontal forces act on the projectile in ideal conditions, its horizontal acceleration is zero ($a_x = 0$). Thus, the horizontal velocity remains constant throughout the entire flight:

   $$v_x = v_0$$

   The horizontal displacement achieved during the time of flight is called the horizontal range ($R$ or $x$):

   $$R = v_x \cdot t = v_0 \cdot t$$

   Substituting the expression for the time of flight ($t = \sqrt{\frac{2h}{g}}$) yields:

   $$R = v_0\sqrt{\frac{2h}{g}}$$

   Therefore, the horizontal range is directly proportional to both the initial launch speed $v_0$ and the square root of the launch height $h$.

   Velocity at Any Instant

   At any time $t$ before impact, the velocity components are:

   • $v_x = v_0$
   • $v_y = gt$

   The magnitude of the instantaneous velocity $v$ is given by the Pythagorean theorem:

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

   The direction of motion, measured as the angle $\theta$ below the horizontal, is:

   $$\tan\theta = \frac{v_y}{v_x} = \frac{gt}{v_0}$$

   Equation of the Trajectory

   Eliminating the parameter $t$ from $x = v_0 t$ and $y = \frac{1}{2}gt^2$ gives the path equation:

   $$y = \frac{g}{2v_0^2}x^2$$

   This quadratic relationship confirms that the path of a horizontally launched projectile is a semi-parabola.

   Application Example

   A stone is thrown horizontally from the top of a cliff of height $45\text{ m}$ with an initial speed of $20\text{ m/s}$. Taking $g = 10\text{ m/s}^2$, determine:

   1. The time of flight.
   2. The horizontal range.
   3. The speed of the stone just before hitting the ground.

   Solution Steps

   1. Time of flight:
   Using $t = \sqrt{\frac{2h}{g}}$:
   $$t = \sqrt{\frac{2 \times 45}{10}} = \sqrt{9} = 3\text{ s}$$

   2. Horizontal range:
   Using $R = v_0 \cdot t$:
   $$R = 20 \times 3 = 60\text{ m}$$

   3. Final impact speed:
   Horizontal velocity remains constant: $v_x = 20\text{ m/s}$.
   Vertical velocity at impact: $v_y = gt = 10 \times 3 = 30\text{ m/s}$.
   Magnitude of resultant velocity:
   $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 30^2} = \sqrt{400 + 900} = \sqrt{1300} \approx 36.1\text{ m/s}$$