JEE Mains Physics · Motion in a Straight Line

Motion Under Gravity

Near the ground every freely moving body has the same downward acceleration g, so the equations of motion apply with a = −g once up is taken as positive, whether the body is dropped, thrown up or thrown down.

All Motion in a Straight Line notes JEE Mains Physics strategy

Why this matters

Twenty-five PYQs, nine of them asking for a number, and four from 2026. Nine drop a body from rest, nine throw one up or down, and seven follow two bodies or a stream of drops at once. Fix the sign convention before writing anything: most wrong options come from a lost minus sign or a lost initial velocity.

Concept 1 of 3: Free fall from rest

A body dropped from rest has u = 0 and gains g in speed every second. The distance fallen grows as t², so it covers most of its fall near the end: the second half of the height takes much less time than the first. To locate a point on the way down, find the speed there with v² = 2gh.

Definition

Taking down as positive: $h = \tfrac{1}{2}gt^{2}$ , $v = gt$ , $v = \sqrt{2gh}$ .
Distance in the nth second: $\dfrac{g}{2}(2n - 1)$ ; with g = 10 that is 5, 15, 25, 35 m.
First and second halves of the height: $t_2 = (\sqrt{2} - 1)\,t_1$ .
The speed equals g in number after 1 s, that is after falling $g/2$ metres.
Two points a known distance apart, passed in a known time: $d = v_At + \tfrac{1}{2}gt^{2}$ gives the speed at the upper point, then $h_A = v_A^{2}/2g$ .
A ball dropped from $h_1$ that rebounds to $h_2$ : it lands at $\sqrt{2gh_1}$ down and leaves at $\sqrt{2gh_2}$ up, so $\Delta v$ is their SUM; average acceleration $= \Delta v/\Delta t$ .
A body moving sideways that drops something: the object falls for $\sqrt{2h/g}$ while the body moves on.

Falling from rest

h = \tfrac{1}{2}gt^{2} \qquad v = gt \qquad v = \sqrt{2gh}

Worked example

A stone is dropped from a cliff 180 m high

(g = 10\ \text{m/s}^{2})

. Find the time of fall, the speed at the bottom and the distance covered in the last second.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q114Moderate

Example 1 · Motion in a Straight Line · Motion Under Gravity

A body falling under gravity covers two points A and B separated by 80 m in 2 s. The distance of upper point

A

from the starting point is _____

m

(use

g = 10{\text{ }ms}^{- 2}

)

A rebound changes the sign of the velocity

Down at 10 m/s and up at 5 m/s is a change of 15 m/s, not 5 m/s. Speeds add when the direction reverses.

Using 9.8 when the stem gives 10

Use the g the question states. The options are usually built on it, and the other value lands near a wrong option.

Drill 8 more on free fall from rest

Concept 2 of 3: Thrown up or down: one equation for the whole flight

With up taken as positive, a = −g for the whole flight: rising, at the top and falling. So one equation, s = ut − ½gt², covers the trip from launch to landing, and a landing point below the start is simply a negative s. A stone released from a rising balloon is not dropped from rest: it starts with the balloon's upward velocity.

Definition

Up positive: $s = ut - \tfrac{1}{2}gt^{2}$ , $v = u - gt$ , $v^{2} = u^{2} - 2gs$ .
Time to the top $u/g$ ; greatest height $u^{2}/2g$ ; it passes the launch level again at speed u, going down.
At the top $v = 0$ , so momentum is zero, but the acceleration is still g, downward.
From a tower: thrown up lands after $t_1$ , thrown down with the same speed after $t_2$ , dropped after $\sqrt{t_1t_2}$ .
The two times at one height are the roots of $h = ut - \tfrac{1}{2}gt^{2}$ ; their product is $2h/g$ .
Released from a body moving at v: the object starts at v, then a = −g.
With a constant air drag f on mass m: $a_{up} = g + f/m$ , $a_{down} = g - f/m$ , and $\dfrac{t_{up}}{t_{down}} = \sqrt{\dfrac{g - f/m}{g + f/m}}$ . The ascent is quicker.

Whole flight, up positive

s = ut - \tfrac{1}{2}gt^{2} \qquad H = \frac{u^{2}}{2g} \qquad t_{drop} = \sqrt{t_1t_2}

Worked example

A ball is thrown up at

20\ \text{m/s}

from the top of a 60 m tower

(g = 10\ \text{m/s}^{2})

. When does it hit the ground, and how high above the ground does it rise?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q4Moderate

Example 2 · Motion in a Straight Line · Motion Under Gravity

A gas balloon is going up with a constant velocity of

10\text{ }m/s

. When this balloon reached a height of 75 m , a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is

\_\_\_\_

m . (Take

g = 10\text{ }m/s^{2}

)

Dropped from a moving body is not dropped from rest

A stone let go from a balloon rising at 8 m/s first rises at 8 m/s. Taking u = 0 gives a shorter fall and the wrong answer.

Zero velocity is not zero acceleration

At the top the body is momentarily at rest, but gravity still acts. If a were zero there, the body would stay at the top.

Splitting the flight when one equation will do

Going up, then down, as two stages is slower and invites sign errors. With a = −g throughout, s = ut − ½gt² handles the whole flight in one line.

Drill 8 more on thrown up or down: one equation for the whole flight

Concept 3 of 3: Two bodies, or drops falling at regular intervals

Write one position equation for each body, using each body's OWN elapsed time, then set the positions equal for a meeting. For drops leaving a tap at equal intervals, count intervals, not drops: when the first lands as the nth leaves, there are n − 1 intervals in one fall time. Because both bodies have the same acceleration g, their relative velocity stays constant.

Definition

Body started $\tau$ later: its elapsed time is $t - \tau$ .
Meeting: equal positions. Two balls thrown up at u, $\tau$ apart, meet at $t = \dfrac{u}{g} + \dfrac{\tau}{2}$ .
Drops: the first lands as the nth begins, so the fall time T splits into $n - 1$ equal intervals; the kth drop has fallen for $T - (k - 1)\dfrac{T}{n - 1}$ .
Juggler throwing n balls a second, each when the last is at the top: time up $= 1/n$ , height $= \dfrac{g}{2n^{2}}$ .
Two bodies in free fall: relative acceleration zero, so their gap changes at a constant rate.

Later body's position

y_2 = u_2(t - \tau) - \tfrac{1}{2}g(t - \tau)^{2}

Worked example

Drops fall from a tap 20 m above the floor at equal intervals. When the first drop hits the floor, the fifth begins to fall. Where are the second and third drops then?

(g = 10\ \text{m/s}^{2})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q1Moderate

Example 3 · Motion in a Straight Line · Motion Under Gravity

Two balls

A

and

B

are placed at the top of

180\text{ }m

tall tower. Ball A is released from the top at

t = 0

s. Ball

B

is thrown vertically down with an initial velocity '

u

' at

t = 2\text{ }s

. After a certain time, both balls meet

100\text{ }m

above the ground. Find the value of '

u

' in

ms^{- 1}

. [use

g = 10{\text{ }ms}^{- 2}

] :

Counting drops instead of intervals

If the first drop lands as the sixth begins, five intervals fit in one fall time, not six.

Giving both bodies the same clock

A body launched later has been moving for t − τ, not t. Using t for both makes them meet at the wrong moment.

Drill 6 more on two bodies, or drops falling at regular intervals

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Free fall from rest
Falling from rest
$h = \tfrac{1}{2}gt^{2} \qquad v = gt \qquad v = \sqrt{2gh}$
Thrown up or down: one equation for the whole flight
Whole flight, up positive
$s = ut - \tfrac{1}{2}gt^{2} \qquad H = \frac{u^{2}}{2g} \qquad t_{drop} = \sqrt{t_1t_2}$
Two bodies, or drops falling at regular intervals
Later body's position
$y_2 = u_2(t - \tau) - \tfrac{1}{2}g(t - \tau)^{2}$

Watch out for (7)

A rebound changes the sign of the velocity
→ Free fall from rest
Using 9.8 when the stem gives 10
→ Free fall from rest
Dropped from a moving body is not dropped from rest
→ Thrown up or down: one equation for the whole flight
Zero velocity is not zero acceleration
→ Thrown up or down: one equation for the whole flight
Splitting the flight when one equation will do
→ Thrown up or down: one equation for the whole flight
Counting drops instead of intervals
→ Two bodies, or drops falling at regular intervals
Giving both bodies the same clock
→ Two bodies, or drops falling at regular intervals

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