AP Physics 1: Free Fall Motion

Free fall motion is a cornerstone concept in kinematics that describes the movement of any object solely under the influence of gravity. Mastering it is essential not only for the AP Physics 1 exam but also for understanding the fundamental principles that govern everything from a falling apple to the trajectory of a rocket. By learning to analyze objects dropped, thrown upward, or thrown downward, you build a critical toolkit for solving a vast array of physics problems.

Defining Free Fall and Gravitational Acceleration

Free fall is defined as any motion of an object where gravity is the only force acting upon it. This means we ignore air resistance and other non-conservative forces. On Earth, this results in a constant downward acceleration, denoted as g. The standard magnitude for this acceleration due to gravity is $g=9.8{\text{ m/s}}^2$. The direction of this acceleration is always toward the center of the Earth, which we typically define as the negative direction on a vertical y-axis.

It is crucial to internalize that acceleration is the rate of change of velocity. In free fall, an object’s velocity changes by 9.8 m/s every second, regardless of whether it is moving upward, downward, or momentarily at rest. For example, if you throw a ball straight up, its upward speed decreases by 9.8 m/s each second until it stops at its peak. Then, as it falls, its downward speed increases by 9.8 m/s each second.

The Surprising Independence of Mass

A classic demonstration involves dropping a heavy bowling ball and a light feather simultaneously in a vacuum. They hit the ground at exactly the same time. Why? Because the acceleration due to gravity is independent of the object's mass. This stems from Newton’s second law ($F=ma$) and the law of universal gravitation. The gravitational force ($F_g=mg$) is proportional to mass. When you set $mg=ma$, the mass $m$ cancels out, leaving $a=g$.

Therefore, all objects in free fall, irrespective of their mass, shape, or composition, accelerate downward at the same rate when air resistance is negligible. This counterintuitive fact is fundamental: a car and a marble dropped from the same height would strike the ground simultaneously in an ideal free-fall scenario.

Applying Kinematic Equations

The motion of an object in free fall is one-dimensional motion with constant acceleration. We use the same suite of kinematic equations, with the acceleration $a$ replaced specifically by $-g$ (if we define upward as positive).

The key equations are: $$v_f=v_0-gt$$ $$y_f=y_0+v_{0}t-\frac{1}{2}g{t}^2$$ $$v_f^2=v_0^2-2g\Delta y$$

Here, $v_0$ is the initial velocity, $v_f$ is the final velocity, $t$ is time, $y_0$ is the initial position, $y_f$ is the final position, and $\Delta y=y_f-y_0$.

Example Problem: A ball is thrown vertically upward from the ground with an initial speed of 20 m/s. How high does it go? Step 1: Identify knowns. At its maximum height, the final velocity $v_f=0$. Initial velocity $v_0=+20\text{ m/s}$, acceleration $a=-g=-9.8{\text{ m/s}}^2$. We want $\Delta y$. Step 2: Choose the right equation. We don’t have time, so we use $v_f^2=v_0^2-2g\Delta y$. Step 3: Solve. $$0^2=(20{)}^2-2(9.8)\Delta y$$ $$0=400-19.6\Delta y$$ $$\Delta y=400/19.6\approx 20.4\text{ meters}$$

Analyzing Symmetry in Trajectory

For an object launched vertically and landing at the same vertical level from which it was launched (like throwing a ball straight up and catching it), the motion exhibits beautiful symmetry. This symmetry provides powerful shortcuts for problem-solving.

The key symmetrical properties are:

1. Time: The time taken to rise to the peak equals the time taken to fall back down from the peak.
2. Speed: At any given height on the way up and on the way down, the object’s speed is the same (though velocities are opposite in direction).
3. Acceleration: The acceleration is constant at $-9.8{\text{ m/s}}^2$ at all points, including the peak where velocity is zero.

Consider the previous example: the ball took approximately 2.04 seconds to reach its peak ($t=v_0/g$). It will take another 2.04 seconds to fall back to the launch point, for a total hang time of about 4.08 seconds. Furthermore, when it returns to your hand, its speed will be 20 m/s, exactly equal to its launch speed (but directed downward).

Problems with an Initial Downward Velocity

Objects thrown downward start with a negative initial velocity (if up is positive). The kinematic equations work identically; you simply plug in a negative value for $v_0$. The acceleration remains $-g$. These problems often ask for the velocity upon impact or the time to fall a certain distance. The process is the same: list knowns, select the equation, and solve, paying careful attention to the signs of displacement, velocity, and acceleration.

Common Pitfalls

1. Ignoring Sign Conventions: The most frequent error is sign inconsistency. You must choose a direction (usually upward) as positive and stick with it for all values: displacement, velocity, and acceleration ($a=-9.8{\text{ m/s}}^2$). Mixing signs leads to incorrect answers.
2. Assuming Acceleration is Zero at the Peak: Velocity is zero at the peak of a trajectory, but acceleration is not. Gravity is still acting, so the acceleration remains $-9.8{\text{ m/s}}^2$. The object is continuously changing its velocity, even at the instant it stops moving upward.
3. Misapplying the "Mass Matters" Intuition: In free fall, mass does not affect acceleration. Do not let everyday experiences with air resistance mislead you into thinking a heavier object falls faster under ideal conditions.
4. Using the Wrong Equation: Students often reach for an equation without checking which variables they know and which they need to find. Always list your knowns ($v_0$, $v_f$, $a$, $\Delta y$, $t$) first to guide your selection of the kinematic equation.

Summary

• Free fall is motion under the influence of gravity alone, characterized by a constant downward acceleration due to gravity, $g=9.8{\text{ m/s}}^2$.
• This acceleration is independent of mass; all objects fall at the same rate when air resistance is negligible.
• Free fall problems are solved using the standard kinematic equations by substituting $a=-g$, requiring strict adherence to a defined sign convention.
• For vertical launches that return to the launch height, the motion exhibits perfect symmetry: time up equals time down, and speed is the same at equal heights.
• Success hinges on consistent sign management and recognizing that acceleration is constant and non-zero at every point, including at the peak of a trajectory where velocity is instantaneously zero.