What is free fall, and why is it so testable?
NCERT Example 2.4: "An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. … If air resistance is neglected, the object is said to be in free fall." The acceleration is the constant $g$, directed downward. For falls small compared with Earth's radius, $g$ stays constant. Free fall is the kinematic equations applied to a vertical axis with $a = \pm g$ fixed.
What makes free fall feel different is psychological, not physical. The same object travels up and down with the same acceleration, and an object momentarily at rest at the top of its flight is famously misread as having zero acceleration. NEET examiners know this and craft distractors around exactly these slips. Free fall has been tested in NEET 2020, 2022 and 2023.
Three facts to anchor:
- $g$ is the same for all objects in free fall — NIOS §2.5 records: "all bodies, irrespective of their size or weight, fall with the same acceleration."
- $g$ acts downward at every instant — during ascent, at the top, during descent. Its sign in your equations depends on the axis; its direction in space does not.
- Free fall is uniformly accelerated motion — $v = v_0 + at$, $x = v_0 t + \tfrac{1}{2}at^2$, $v^2 = v_0^2 + 2a(x-x_0)$ all apply with $a = \pm g$.
Sign convention and the value of $g$
The first rule of vertical-motion problems is state the axis, then never change it. NCERT Points to Ponder 1: "The origin and the positive direction of an axis are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration."
The recommended convention — and the one most NCERT examples use — is upward positive, with the origin at the point of launch. Then $a = -g$; an upward throw has $u > 0$; a drop has $u = 0$; a downward throw has $u < 0$. The opposite choice (downward positive, $a = +g$) gives identical physical answers.
For numerics, use $g = 10\text{ m s}^{-2}$ as the default NEET value. The true magnitude is about $9.8\text{ m s}^{-2}$, and NCERT Example 2.4 uses that, but recent NEET papers — including NEET 2023 Q.40 and NEET 2020 Q.114 — specify $g = 10$ to keep arithmetic clean. Unless told otherwise, use $g = 10$.
The three canonical scenarios
Every free-fall NEET question — and there are many of them — collapses into one of three set-ups. Memorise the formulae once; recognise the scenario fast; substitute. The table below summarises all three with upward positive, origin at the launch point.
| Scenario | Initial conditions | Time of flight | Max height above launch | Impact / return speed |
|---|---|---|---|---|
| A. Dropped from rest from height $h$ | $u = 0$, $a = -g$, lands at $y = -h$ | $t = \sqrt{2h/g}$ | (rises to 0; falls $h$) | $v = \sqrt{2gh}$ (downward) |
| B. Thrown upward with speed $u$ from ground | $u > 0$, $a = -g$, returns to $y = 0$ | Total $T = 2u/g$; up-time $= u/g$ | $H = u^2 / (2g)$ | $|v| = u$ on return (speed-symmetry) |
| C. Thrown upward with speed $u$ from height $H_0$ | $u > 0$ (or $<0$ if downward), $a = -g$, lands at $y = -H_0$ | Solve $-H_0 = ut - \tfrac{1}{2}gt^2$ quadratic | Rises $u^2/(2g)$ above launch (if $u > 0$) | $v = -\sqrt{u^2 + 2gH_0}$ (magnitude) |
Scenario A — Dropped from rest from a height
The simplest case. An object is released (not thrown) from height $h$ above the ground. Take origin at the release point, upward positive: $u = 0$, $a = -g = -10\text{ m s}^{-2}$. The position $y$ becomes negative as the object falls; it hits the ground when $y = -h$.
The three workhorse equations specialise to:
$$y = -\tfrac{1}{2} g t^2 \qquad v = -g t \qquad v^2 = 2g\,|y|$$From the first: setting $|y| = h$, the time to hit the ground is
$$t = \sqrt{\frac{2h}{g}}.$$From the third: the impact speed is
$$|v| = \sqrt{2 g h}.$$A ball is dropped from rest from a height of 80 m. Find (a) the time to reach the ground and (b) the speed with which it strikes the ground. Take $g = 10\text{ m s}^{-2}$.
Setup. Origin at release point, upward positive. $u = 0$, $a = -g = -10\text{ m s}^{-2}$, ground at $y = -80\text{ m}$.
(a) Time. $y = -\tfrac{1}{2}gt^2 \Rightarrow -80 = -\tfrac{1}{2}(10) t^2 \Rightarrow t^2 = 16 \Rightarrow t = 4\text{ s}$.
(b) Impact speed. $v = -gt = -(10)(4) = -40\text{ m s}^{-1}$. The negative sign means downward; the speed (magnitude) is $40\text{ m s}^{-1}$.
Cross-check using $v^2 = 2gh$: $v^2 = 2(10)(80) = 1600 \Rightarrow |v| = 40\text{ m s}^{-1}$. Consistent.
Same problem with the opposite sign convention
To prove the answer does not depend on the axis, redo with downward positive, origin at release. Now $a = +g = +10$, and the ground is at $y = +80\text{ m}$.
$$y = \tfrac{1}{2}gt^2 \Rightarrow 80 = \tfrac{1}{2}(10) t^2 \Rightarrow t = 4\text{ s}, \qquad v = gt = +40\text{ m s}^{-1}.$$Identical $t$, identical $|v|$. The sign of $v$ flipped because "positive" now means downward. NCERT Points to Ponder 5: equations are algebraic; they work in all situations provided signs are consistent.
Scenario B — Thrown vertically upward from the ground
A ball leaves the hand at ground level with upward speed $u$, decelerates against gravity, momentarily stops at the highest point, and returns to the hand at the same level with the same speed $u$ but in the opposite direction. Upward positive, $a = -g$:
$$v(t) = u - gt \qquad y(t) = ut - \tfrac{1}{2}g t^2 \qquad v^2 = u^2 - 2gy.$$At the highest point, $v = 0$:
$$0 = u - g\,t_\text{up} \Rightarrow t_\text{up} = \frac{u}{g}.$$Maximum height $H$ — the value of $y$ when $v = 0$ — comes from $v^2 = u^2 - 2gy$:
$$0 = u^2 - 2gH \Rightarrow H = \frac{u^2}{2g}.$$To find the total time of flight $T$ — the time to return to $y = 0$ — set $y = 0$ in the position equation:
$$0 = uT - \tfrac{1}{2}gT^2 \Rightarrow T(u - \tfrac{1}{2}gT) = 0 \Rightarrow T = \frac{2u}{g}.$$Notice $T = 2 t_\text{up}$: the time up equals the time down, because launch and landing levels coincide. The return speed at $y = 0$ is
$$v(T) = u - g \cdot \frac{2u}{g} = -u,$$negative because the ball is now moving downward; the speed on return is the same $u$ it left with.
NCERT Exercise 2.6 (paraphrased). A player throws a ball upwards with an initial speed of $29.4\text{ m s}^{-1}$. To what height does the ball rise, and after how long does it return to the player's hands? (Take $g = 9.8\text{ m s}^{-2}$, neglect air resistance.)
Max height. $H = \dfrac{u^2}{2g} = \dfrac{(29.4)^2}{2(9.8)} = \dfrac{864.36}{19.6} = 44.1\text{ m}$.
Total time of flight. $T = \dfrac{2u}{g} = \dfrac{2 \times 29.4}{9.8} = 6\text{ s}$.
NCERT chose $u = 29.4\text{ m s}^{-1}$ because it is exactly $3g$ (with $g = 9.8$), making the numbers fall out cleanly.
A boy throws a stone vertically upward from the ground with speed $20\text{ m s}^{-1}$. With $g = 10\text{ m s}^{-2}$, find (a) the maximum height reached, (b) the total time of flight, and (c) the velocity of the stone when it returns to his hand.
(a) Max height. $H = \dfrac{u^2}{2g} = \dfrac{(20)^2}{2(10)} = \dfrac{400}{20} = 20\text{ m}$.
(b) Time of flight. $T = \dfrac{2u}{g} = \dfrac{2 \times 20}{10} = 4\text{ s}$.
(c) Return velocity. $v = u - gT = 20 - 10 \times 4 = -20\text{ m s}^{-1}$ — same speed, opposite direction.
Scenario C — Thrown vertically from a height
The NEET-favourite scenario. An object is launched from a tower of height $H_0$ with initial vertical velocity $u$ (upward, downward, or zero) and lands at ground level. Origin at launch, upward positive, ground at $y = -H_0$.
The position equation is the workhorse:
$$y = u t - \tfrac{1}{2} g t^2.$$Setting $y = -H_0$ and solving for $t$:
$$-H_0 = u t - \tfrac{1}{2}gt^2 \quad\Longrightarrow\quad \tfrac{1}{2}g t^2 - u t - H_0 = 0.$$This is a quadratic in $t$. The physical (positive) root gives the total time from launch to ground impact. The impact speed follows from
$$v^2 = u^2 + 2 g H_0 \qquad |v| = \sqrt{u^2 + 2gH_0}.$$A ball is thrown vertically upwards with a velocity of $20\text{ m s}^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0\text{ m}$ from the ground. (a) How high will the ball rise? (b) How long will it be before the ball hits the ground? Take $g = 10\text{ m s}^{-2}$.
Setup. Origin at launch (rooftop), upward positive. $u = +20$, $a = -10$, ground at $y = -25$.
(a) Max height above launch. $H = u^2/(2g) = 400/20 = 20\text{ m}$. The ball reaches $20\text{ m}$ above the rooftop, i.e., $45\text{ m}$ above the ground.
(b) Total time to ground. $-25 = 20 t - 5 t^2 \Rightarrow 5t^2 - 20t - 25 = 0 \Rightarrow t^2 - 4t - 5 = 0 \Rightarrow (t-5)(t+1) = 0 \Rightarrow t = 5\text{ s}$ (the negative root is unphysical).
Sanity check by splitting. Time to reach max height: $t_\text{up} = u/g = 2\text{ s}$. From the max height ($45\text{ m}$ above ground) the ball is in free fall from rest; $45 = \tfrac{1}{2}(10)t_\text{down}^2 \Rightarrow t_\text{down} = 3\text{ s}$. Total $= 2 + 3 = 5\text{ s}$. Matches.
A student standing on a bridge throws a small ball vertically upwards with a velocity $4\text{ m s}^{-1}$. The ball strikes the water surface after $4$ s. Find the height of the bridge above the water surface. Take $g = 10\text{ m s}^{-2}$.
Setup. Origin at the bridge, upward positive. $u = +4$, $a = -10$, $t = 4\text{ s}$. We want $|y(4)|$.
$y(4) = u t - \tfrac{1}{2}g t^2 = 4(4) - \tfrac{1}{2}(10)(4)^2 = 16 - 80 = -64\text{ m}$.
So the water surface is $64\text{ m}$ below the bridge. Height of bridge $= 64\text{ m}$. The distractor $68\text{ m}$ comes from forgetting the $u t$ term (treating the throw as a drop); $56\text{ m}$ from sign-flipping the $ut$ contribution.
A ball is thrown vertically downward with a velocity of $20\text{ m s}^{-1}$ from the top of a tower. It hits the ground with velocity $80\text{ m s}^{-1}$. Find the height of the tower, with $g = 10\text{ m s}^{-2}$.
Setup. Downward positive (matching the direction of motion to keep signs simple). $u = +20$, $a = +g = +10$, $v = +80$. Use $v^2 = u^2 + 2 a h$.
$80^2 = 20^2 + 2(10) h \Rightarrow 6400 = 400 + 20 h \Rightarrow h = 300\text{ m}$.
The same problem with upward positive gives $u = -20$, $v = -80$, $a = -10$; $v^2 - u^2 = 6400 - 400 = 6000$, and $2 a h = -20 h$ with $h$ here being the displacement, which is $-300\text{ m}$ (the ball ends up 300 m below the start). Same magnitude either way.
$n$th-second formula and Galileo's odd numbers
NCERT Example 2.5 reports Galileo's law of odd numbers: "The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity (namely, 1 : 3 : 5 : 7 …)." NEET 2022 Q.5 tested this directly. The result follows from the general $n$th-second formula for uniformly accelerated motion:
$$s_n = u + \frac{a}{2}(2n - 1).$$For a body dropped from rest, $u = 0$ and $a = g$, so the distance covered during the $n$th second is
$$s_n = \frac{g}{2}(2n - 1).$$With $g = 10\text{ m s}^{-2}$, this gives
- $s_1 = 5\text{ m}$ (the first second)
- $s_2 = 15\text{ m}$ (the second second)
- $s_3 = 25\text{ m}$ (the third second)
- $s_4 = 35\text{ m}$ (the fourth second)
- $s_5 = 45\text{ m}$ (the fifth second)
Ratio $1 : 3 : 5 : 7 : 9$, just as Galileo claimed. Note carefully: $s_n$ is the distance covered during the $n$th second of fall — the interval from $t = n-1$ to $t = n$ — not the total distance covered in $n$ seconds. The total distance after $n$ seconds is $\tfrac{1}{2}g n^2$, not $\tfrac{g}{2}(2n - 1)$.
NCERT Exercise 2.8 — a ball dropped on the floor losing one-tenth of its speed at every collision — applies Scenario A to a sequence of progressively shorter falls; each bounce still obeys $v = \sqrt{2gh}$ and $t = \sqrt{2h/g}$.
What this subtopic locked in
- Free fall is uniformly accelerated motion with $a = g$ downward. All kinematic equations apply.
- $g = 10\text{ m s}^{-2}$ is the NEET default. Use 9.8 only if the question says so.
- State the axis first. Upward-positive is recommended; sign of $a$ follows the choice.
- At the highest point: $v = 0$, $a = g$ downward. Velocity zero $\ne$ acceleration zero.
- Time up = time down only when launch and landing are at the same level. Otherwise solve the quadratic.
- Drop from rest: $t = \sqrt{2h/g}$, $v = \sqrt{2gh}$.
- Throw upward from ground: $H = u^2/(2g)$, $T = 2u/g$, return speed $= u$.
- $n$th-second formula: $s_n = \tfrac{g}{2}(2n-1)$ for free fall from rest. Ratio is $1 : 3 : 5 : 7$.