Free Fall and Motion Under Gravity

Every vertical motion near the Earth's surface — a dropped stone, a tossed ball, a diver leaving the board — runs on one number: g = 9.8 m s⁻², directed downward. Because that acceleration is constant, the kinematic equations apply directly, and free fall becomes the cleanest possible application of the whole chapter. It is also one of NTA's most reliable question factories: the "v = 0 but a = 9.8 m s⁻²" highest-point trap, maximum-height and time-of-flight numericals, up–down symmetry checks, and the reaction-time ruler experiment recycled as an integer-answer question. This page assembles the full toolkit — dropped bodies, bodies thrown up, bodies thrown down — with a strict sign convention so that no sign marks are ever lost, and closes with the classic experiments that made g measurable in the first place.

What is Free Fall? — Complete Theory

1. The model and its three conditions

Free fall is motion under gravity alone: air resistance is neglected, the body stays near the Earth's surface where g does not vary with height, and the path is a vertical straight line. Under those three conditions every falling body — feather or iron ball — has the same acceleration g = 9.8 m s⁻² downward, a fact that shocked Aristotle's followers and still shocks students: mass simply never enters the kinematics. The value 9.8 m s⁻² is itself an acceleration (dimension [L T⁻²]), so writing "a force of 9.8" or "g newtons" is a dimensional error — g in newtons belongs to W = mg, a different relation altogether (Chapter 4). All formulas below are the kinematic equations with a replaced by g; they inherit every validity condition of those equations.

2. Body dropped from rest (take down as positive here)

With v₀ = 0 and a = +g (down = positive for a pure drop — the cleanest convention):

v = gt · h = ½gt² · v = √(2gh)

The third relation — obtained by eliminating t — says the impact speed depends only on the drop height, not on the mass or the time taken. Quick calibration numbers worth memorising: after 1 s, v = 9.8 m s⁻¹ and h = 4.9 m; after 2 s, v = 19.6 m s⁻¹ and h = 19.6 m. Doubling the time quadruples the distance (t² law) — a proportionality NTA tests directly: "a body falls n times farther in the nth second than the first?" style items reduce to the odd-number ladder 1 : 3 : 5 : … from the kinematic page.

3. Body thrown upward (now take up as positive, so a = −g)

With initial speed u upward:

v = u − gt · h = ut − ½gt² · at the top v = 0

Setting v = 0 gives the two headline results:

t_up = u/g (time to the top) · h_max = u²/2g (maximum height)

The descent is the ascent played backwards: with no air drag the body regains the launch point with speed u (speed symmetry), after a total flight time T = 2u/g — time up equals time down. The full v–t graph is a single straight line of slope −g crossing zero exactly once, at the top; nothing about the graph is "sharper" at the crossing, which is precisely why v = 0 there does not imply a = 0. If the body is thrown downward with speed u₀ instead, simply use v = u₀ + gt with down positive — one convention, applied without apology, is the whole skill.

4. Reaction time: g as a stopwatch (NCERT experiment)

Have a friend hold a ruler vertically and drop it without warning; catch it and read the drop distance d. Your brain's delay is t = √(2d/g), straight from h = ½gt². For d = 19.6 cm = 0.196 m: t = √(2 × 0.196/9.8) = √0.04 = 0.2 s — a typical human reaction time, which is why this example survives in every exam cycle: it converts a length measurement into a time measurement using nothing but g.

Visualising a Vertical Throw: One Line, Whole Story

Solved Examples on Free Fall (Step-by-Step)

Solved Example 1 — Dropped from 19.6 m

Q. A stone is dropped from a cliff 19.6 m high. Find the time to reach the ground and the impact speed (g = 9.8 m s⁻²).

Step 1. Take down positive, v₀ = 0. Time: h = ½gt² → 19.6 = ½(9.8)t² = 4.9t² → t² = 4 → t = 2 s.

Step 2. Speed: v = gt = 9.8 × 2 = 19.6 m s⁻¹.

Step 3. Cross-check with the time-free form: v = √(2gh) = √(2 × 9.8 × 19.6) = √384.16 = 19.6 m s⁻¹ ✔.

Solved Example 2 — Thrown up at 19.6 m s⁻¹

Q. A ball is thrown vertically upward at 19.6 m s⁻¹. Find the maximum height, the time to the top, and the total time of flight.

Step 1. Take up positive, a = −g. Top: v = 0 → t_up = u/g = 19.6/9.8 = 2 s.

Step 2. Height: h_max = u²/2g = (19.6)²/(2 × 9.8) = 384.16/19.6 = 19.6 m.

Step 3. Flight: T = 2u/g = 2 × 2 = 4 s; the ball returns to the hand at 19.6 m s⁻¹ downward (symmetry).

Practice Questions on Free Fall (With Solutions)

Practice Q1 (Numerical-value type)

Q. A body falls from rest and strikes the ground at 49 m s⁻¹. From what height did it fall? (g = 9.8 m s⁻²; answer in metres)

Solution. v² = 2gh → h = v²/2g = (49)²/(2 × 9.8) = 2401/19.6 = 122.5 m. Mass and fall time never entered the calculation.

Practice Q2 (Single-correct MCQ)

Q. A stone is dropped from rest (down positive). Its speed after 3 s is: (a) 9.8 m s⁻¹ (b) 19.6 m s⁻¹ (c) 29.4 m s⁻¹ (d) 44.1 m s⁻¹

Solution. v = gt = 9.8 × 3 = 29.4 m s⁻¹. (Distance fallen in that time is 44.1 m = ½ × 9.8 × 9 — option (d) is the planted distance-for-speed swap.)

Practice Q3 (Assertion–Reason)

Q. Assertion (A): In vacuum, a coin and a feather released together from the same height reach the floor at the same instant. Reason (R): The acceleration of free fall is independent of the body's mass.
Options: (a) both true, R explains A · (b) both true, R does not explain A · (c) A true, R false · (d) A false, R true

Solution. With drag absent, both bodies share a = g regardless of mass, so equal heights take equal times. R is the physical reason A holds.

Key Formulas & Takeaways

FAQs on Free Fall and Motion Under Gravity

What is free fall?

Motion under gravity alone, with air resistance neglected. The acceleration is g = 9.8 m s⁻² directed downward for every body, independent of its mass.

What is the maximum height of a ball thrown up with speed u?

h_max = u²/2g, reached after t_up = u/g. For u = 19.6 m s⁻¹ and g = 9.8 m s⁻², the height is 19.6 m after 2 s.

What is the acceleration at the highest point of a vertical throw?

9.8 m s⁻² downward — full gravity — even though the velocity is momentarily zero there. Zero velocity does not mean zero acceleration.

Do heavier objects fall faster?

No. In free fall without air resistance the acceleration g is the same for all masses, so a stone and a feather released together strike the ground together in vacuum.