AP Physics 1: Horizontal Projectile Motion

Understanding horizontal projectile motion is essential because it provides the clearest model for how two-dimensional motion works. This concept is a cornerstone of mechanics and is heavily tested on the AP Physics 1 exam. Mastering it will allow you to analyze everything from a car driving off a cliff to a supply package dropped from a plane.

The Core Principle: Independence of Motion

The foundation of all projectile motion analysis is the independence of motion in perpendicular directions. This means the horizontal and vertical components of a projectile's motion are completely separate and do not influence each other. The only link between them is the variable time.

Imagine dropping a coin straight down from rest at the exact same moment you flick a second coin horizontally off the table. Both coins will hit the ground simultaneously. The horizontal motion given to the second coin does nothing to speed up or slow down its vertical fall under gravity. This principle allows us to break a complex two-dimensional problem into two simpler, one-dimensional problems: constant velocity in the horizontal direction and constant acceleration ($a_y=g=-9.8{\text{ m/s}}^2$) in the vertical direction.

Component #1: Vertical Free Fall (Solving for Time)

For a projectile launched horizontally, its initial vertical velocity is always zero ($v_{iy}=0$). Its vertical motion is identical to an object in free fall dropped from rest. We use the kinematics equations for constant acceleration to analyze this dimension.

The most critical equation here is: $$\Delta y=v_{iy}t+\frac{1}{2}{a}_y{t}^2$$ Since $v_{iy}=0$ and $a_y=g$, this simplifies to: $$\Delta y=\frac{1}{2}g{t}^2$$ Remember that $\Delta y$ is the vertical displacement, which for a projectile launched from a height is negative (if we define upward as positive). The magnitude is the launch height. Solving for time of flight ($t$): $$t=\sqrt{\frac{2|\Delta y|}{g}}$$ This time of flight depends only on the launch height and the acceleration due to gravity. The initial horizontal speed has no effect on how long the object is in the air.

Component #2: Horizontal Constant Velocity (Solving for Range)

In the absence of air resistance, there is no horizontal acceleration ($a_x=0$). Therefore, the horizontal component of velocity remains constant: $v_x=v_{ix}=\text{constant}$.

The horizontal displacement, or range ($R$), is found using the constant velocity equation: $$\Delta x=v_{x}t$$ Since the horizontal velocity is constant, this simplifies to: $$R=v_{ix}t$$ You must use the same time ($t$) calculated from the vertical motion. This is the crucial step that connects the two independent components. Doubling the horizontal launch velocity will double the range, but it will not change the time of flight.

The Finale: Vector Addition for Final Velocity

The final velocity just before impact is a vector with both horizontal and vertical components. The horizontal component $v_{fx}$ remains unchanged from the launch: $v_{fx}=v_{ix}$.

The final vertical component $v_{fy}$ is found using a vertical kinematics equation (with $v_{iy}=0$): $$v_{fy}=v_{iy}+a_{y}t\to v_{fy}=gt$$ Note that $v_{fy}$ will be negative (downward). To find the magnitude of the final velocity vector, use the Pythagorean theorem: $$v_f=\sqrt{(v_{fx}{)}^2+(v_{fy}{)}^2}$$ To find the direction (the angle $\theta$ below the horizontal), use the inverse tangent function: $$\theta ={\tan }^{-1}\left(\frac{|v_{fy}|}{v_{fx}}\right)$$ The final velocity is always directed at an angle downward, and its magnitude is always greater than the initial horizontal launch velocity because the vertical component adds to it.

A Complete Worked Example

Scenario: A ball is thrown horizontally from the roof of a 45.0 m tall building with an initial speed of 12.0 m/s. Find (a) its time of flight, (b) its range, and (c) its final velocity magnitude and direction just before hitting the ground.

Step-by-Step Solution:

1. Define Coordinates: Up is positive. Initial position: $y_i=0$, $x_i=0$. Final: $y_f=-45.0\text{ m}$, $x_f=R$.
2. Knowns: $v_{ix}=12.0\text{ m/s}$, $v_{iy}=0$, $a_y=-9.8{\text{ m/s}}^2$, $\Delta y=-45.0\text{ m}$.

(a) Time of Flight (from vertical motion): $$\Delta y=\frac{1}{2}{a}_y{t}^2$$ $$-45.0=\frac{1}{2}(-9.8)t^2$$ $$t^2=\frac{2\times 45.0}{9.8}=9.184$$ $$t=\sqrt{9.184}=3.03\text{ s}$$

(b) Range (from horizontal motion): $$R=v_{ix}\cdot t=(12.0\text{ m/s})(3.03\text{ s})=36.4\text{ m}$$

(c) Final Velocity: Horizontal Component: $v_{fx}=v_{ix}=12.0\text{ m/s}$ Vertical Component: $v_{fy}=a_{y}t=(-9.8{\text{ m/s}}^2)(3.03\text{ s})=-29.7\text{ m/s}$ Magnitude: $v_f=\sqrt{(12.0{)}^2+(-29.7{)}^2}=\sqrt{144+882}=\sqrt{1026}=32.0\text{ m/s}$ Direction: $\theta ={\tan }^{-1}\left(\frac{29.7}{12.0}\right)={\tan }^{-1}(2.48)=68.0^{\circ }$ below the horizontal.

Common Pitfalls

1. Mixing Horizontal and Vertical Variables in a Single Equation: The most frequent and critical error. You cannot plug vertical displacement into an equation to solve for horizontal velocity. Always identify which component (x or y) each variable belongs to before writing an equation. Correction: Solve for time using a vertical equation, then use that time in a horizontal equation.

1. Sign Errors with Gravity and Displacement: Forgetting that acceleration due to gravity ($a_y=g$) is negative in a standard coordinate system (where up is positive) leads to incorrect times and velocities. Similarly, the vertical displacement ($\Delta y$) for an object falling from a height is negative. Correction: Consistently define your coordinate system at the start and stick to the signs.

1. Assuming the Final Vertical Velocity is Zero: The object is in free fall until it hits the ground; it does not slow down vertically. The final vertical velocity is at its maximum magnitude. Correction: Use $v_{fy}=gt$ (with correct sign), not zero.

1. Confusing Speed with Velocity: The question may ask for the final speed (a scalar) or final velocity (a vector with magnitude and direction). Correction: Read questions carefully. "Speed" requires only the magnitude $v_f$. "Velocity" requires both the magnitude and the direction angle.

Summary

• Horizontal and vertical motions are independent. Time is the only common variable. The vertical motion dictates how long the projectile is in the air, while the horizontal motion dictates how far it goes during that time.
• The time of flight is determined solely by the vertical free fall from the launch height. It is calculated using $\Delta y=\frac{1}{2}g{t}^2$.
• The range is the product of the constant horizontal velocity and the time of flight: $R = v_{ix}