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Worked Examples · Example 2.4

Q.Free-fall: Discuss the motion of an object under free fall. Neglect air resistance.

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Free fall is the motion of an object under the sole influence of gravity, with no air resistance. The object accelerates downward at g≈9.8 m/s2g \approx 9.8 \, \text{m/s}^2, and its motion is described by the equations of uniformly accelerated motion. The key result: all objects in free fall near Earth's surface have the same acceleration, regardless of mass.

Concept and Intuition

Free fall is a special case of projectile motion where the only force acting on an object is gravity. When we neglect air resistance, the object experiences a constant downward acceleration of magnitude gg (approximately 9.8 m/s29.8 \, \text{m/s}^2 near Earth's surface). This is a beautiful simplification: the motion becomes uniformly accelerated motion in the vertical direction.

The most famous demonstration of this principle is Galileo's Leaning Tower of Pisa experiment (whether apocryphal or not) — a heavy ball and a light ball dropped from the same height hit the ground at the same time. This is because gravitational acceleration is independent of mass. In a vacuum, a feather and a hammer fall identically, as shown on the Moon during the Apollo 15 mission.

For free fall, we typically take the downward direction as positive (or upward as positive — be consistent). The equations of motion become:

  • v=u+gtv = u + gt (velocity after time tt)
  • s=ut+12gt2s = ut + \frac{1}{2}gt^2 (displacement after time tt)
  • v2=u2+2gsv^2 = u^2 + 2gs (relation between velocity and displacement)

where uu is initial velocity, vv is final velocity, ss is displacement, and tt is time.

Watch out

A common mistake is forgetting that the sign convention matters. If you take upward as positive, then gg becomes −9.8 m/s2-9.8 \, \text{m/s}^2. Always state your sign convention clearly before solving.

Step-by-Step Analysis

1. Define the sign convention and parameters.

Let's take the downward direction as positive. This means:

  • Acceleration a=+g=9.8 m/s2a = +g = 9.8 \, \text{m/s}^2
  • If the object is dropped from rest, initial velocity u=0u = 0
  • If thrown downward, uu is positive
  • If thrown upward, uu is negative (since upward is opposite to our positive direction)

2. Consider the case of an object dropped from rest.

This is the simplest free-fall scenario. The object starts with zero initial velocity and falls straight down.

Using s=ut+12gt2s = ut + \frac{1}{2}gt^2 with u=0u = 0:

s=12gt2s = \frac{1}{2}gt^2

The distance fallen is proportional to the square of the time. After 1 second, the object falls about 4.9 m4.9 \, \text{m}; after 2 seconds, about 19.6 m19.6 \, \text{m}; after 3 seconds, about 44.1 m44.1 \, \text{m}.

The velocity after time tt is:

v=gtv = gt

So after 1 second, v=9.8 m/sv = 9.8 \, \text{m/s}; after 2 seconds, v=19.6 m/sv = 19.6 \, \text{m/s}; and so on. The velocity increases linearly with time.

3. Consider an object thrown upward.

When you throw an object upward, it initially moves against gravity. Taking downward as positive, the initial velocity uu is negative. The equations become:

v=−u+gt(careful with signs)v = -u + gt \quad \text{(careful with signs)}

Better to use the vector form: v=u+atv = u + at where uu is negative and a=+ga = +g.

The object rises, slows down, reaches a maximum height where v=0v = 0, then falls back down. The time to reach maximum height is found from v=0v = 0:

0=u+gt  ⟹  t=−ug0 = u + gt \implies t = -\frac{u}{g}

Since uu is negative, tt is positive — which makes sense.

Tip

For an object thrown upward, the time to go up equals the time to come back down to the same height (symmetry of motion under constant acceleration). The speed when it returns equals the initial speed, but in the opposite direction.

4. The independence of mass.

This is the most counterintuitive and important result. All objects in free fall — regardless of mass, size, or shape (in vacuum) — accelerate at the same rate gg. Why?

From Newton's second law: F=maF = ma. The gravitational force on an object is F=mgF = mg (where mm is the gravitational mass). Setting these equal:

mg=ma  ⟹  a=gmg = ma \implies a = g

The mass cancels out! This is because gravitational mass and inertial mass are equivalent (the equivalence principle, which Einstein used as the foundation of general relativity).

a=g(independent of mass)a = g \quad \text{(independent of mass)}

5. Real-world considerations.

In practice, air resistance complicates free fall. A feather falls slower than a hammer because air drag opposes motion. The drag force depends on shape, cross-sectional area, and speed. For dense, streamlined objects (like a metal ball), air resistance is negligible over short distances, so free-fall equations work well.

Terminal velocity occurs when air drag equals weight, giving zero net acceleration. For a skydiver, terminal velocity is about 55 m/s55 \, \text{m/s} (200 km/h) in a spread-eagle position, and higher when diving. …

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