Motion of objects under gravity

Drop a ball, throw it straight up, or watch a coin fall off a table — every one of these motions obeys the exact same three equations of motion you already learned for a car speeding up on a straight road. The only change is that the acceleration is no longer some arbitrary $a$ ; it is $g$ , the fixed downward pull of the Earth, worth about $9.8\ \text{m/s}^2$ near the surface. This section takes the three equations of motion from the Motion chapter, swaps in $g$ for $a$ and $h$ for $s$ , and shows how a careful choice of sign convention lets you handle a dropped stone and a ball thrown skyward with the very same formulas. Along the way you will see why an object thrown up momentarily stops but never stops accelerating, and why the trip up takes exactly as long as the trip down.

The core insight

Free fall and vertical throws are uniformly accelerated motion with $a = g \approx 9.8\ \text{m/s}^2$ directed downward, so $v = u+gt$ , $h = ut+\tfrac12gt^2$ , and $v^2=u^2+2gh$ apply directly once a consistent sign convention fixes whether up or down is positive.

The core idea

From $a$ to $g$ . The three equations of motion for any object moving with constant acceleration are

v = u + at, \qquad s = ut + \tfrac12 at^2, \qquad v^2 = u^2 + 2as

where: $u$ is initial velocity, $v$ is final velocity, $a$ is the (constant) acceleration, $t$ is time elapsed, and $s$ is the distance covered. For any object moving purely vertically under gravity — dropped, thrown up, or thrown down — the acceleration is constant, equal to $g$ , and always points toward the Earth's centre. Substituting $a \to g$ and $s \to h$ (height) gives the working forms used throughout this topic:

v = u + gt, \qquad h = ut + \tfrac12 gt^2, \qquad v^2 = u^2 + 2gh

where: $g \approx 9.8\ \text{m/s}^2$ is the acceleration due to gravity and $h$ is the vertical distance covered. No new physics is introduced here — it is the same three equations, applied to a special, very common case of constant acceleration.

Sign convention is everything. Gravity always pulls downward, but the direction you call "positive" depends on the problem, and every quantity in that problem — $u$ , $v$ , $g$ , $h$ — must follow the same choice.

Dropping / falling downward: take downward as positive. Then $u=0$ if released from rest, $g=+9.8\ \text{m/s}^2$ , and $h, v$ both come out positive and growing, since the object speeds up as it falls.
Thrown straight upward: take upward as positive. Then the launch speed $u$ is positive, but $g=-9.8\ \text{m/s}^2$ , because gravity opposes the upward motion. The object decelerates, its velocity passes through zero at the top, and it then "accelerates" back down — all still governed by the same constant $g$ .

Symmetry of the up-and-down trip. Ignoring air resistance, an object thrown straight up returns to its launch point with the same speed it left with, and the time taken to climb to the top exactly equals the time taken to fall back down to the same level. This halves the work in many problems: solve for the time or speed on the way up, and the way down mirrors it.

Figure

A dropped ball speeding up on the way down beside a thrown ball slowing to a stop then speeding up again on the way back down, with velocity arrows growing and shrinking to show where g points at each instant

Derivation AGetting the "equations of motion under gravity" from the general equations of motion

Starting from the general uniformly-accelerated-motion equations $v = u+at$ , $s = ut+\tfrac12at^2$ , $v^2=u^2+2as$ , show precisely how they specialise to a vertically falling or vertically thrown object once $a$ is replaced by the constant, always-downward acceleration $g$ and the distance $s$ is replaced by the vertical height $h$ , and explain what stays true and what changes when the object moves upward instead of downward.

Step 1. Begin with the first equation of motion, $v = u + at$ . It holds for any motion where the acceleration is constant in size and direction over the interval considered — the derivation (from the definition $a = \frac{v-u}{t}$ , rearranged) never assumed the acceleration was due to a car engine, a falling weight, or anything else. It only needed $a$ to be constant.

Step 2. For an object moving purely vertically near the Earth's surface, air resistance neglected, the only force acting is gravity, and it produces a constant acceleration of magnitude $g \approx 9.8\ \text{m/s}^2$ directed toward the Earth's centre — that is, straight down, regardless of whether the object happens to be moving up, down, or momentarily at rest. This lets you write $a = g$ directly in each equation of motion.

Step 3. The displacement in these vertical problems is conventionally called $h$ (height) rather than $s$ , purely as labelling — it plays exactly the same role $s$ played in the general equations. Substituting gives $v = u+gt$ , $h = ut+\tfrac12gt^2$ , and $v^2 = u^2+2gh$ , with no change to the underlying algebra.

Step 4. The one genuinely new ingredient is the sign of $g$ , which depends on which direction you have chosen as positive for that problem. If downward is positive (natural for a dropped or downward-thrown object), $g$ is entered as $+9.8\ \text{m/s}^2$ . If upward is positive (natural for an object thrown up, since its initial velocity $u$ is then a clean positive number), $g$ must be entered as $-9.8\ \text{m/s}^2$ , since gravity opposes the chosen positive direction. Every other quantity in the same problem ( $u$ , $v$ , $h$ ) must use that same sign convention consistently, or the equations give the wrong answer.

Step 5. A useful check on Step 4: at the topmost point of a vertical throw, setting $v=0$ in $v=u+gt$ with $g$ negative gives a positive, sensible time to reach the top — confirming the sign choice was applied correctly, since a physically real "time to reach maximum height" cannot come out negative for an object launched upward.

Worked example

A ball is dropped from rest from the top of a tower 44.1 m high, and separately, a stone is thrown straight up from the ground with an initial speed of 19.6 m/s. Take $g = 9.8\ \text{m/s}^2$ throughout, and find (a) the time and speed with which the dropped ball hits the ground, and (b) the maximum height the thrown stone reaches and the total time for it to return to the thrower's hand.

Step 1. For the dropped ball, take downward as positive, so $u=0$ , $h=44.1\ \text{m}$ , $g=+9.8\ \text{m/s}^2$ . Using $h=ut+\tfrac12gt^2$ with $u=0$ :

44.1 = \tfrac12 \times 9.8 \times t^2 = 4.9\,t^2 \implies t^2 = \frac{44.1}{4.9}=9 \implies t = 3\ \text{s}

Step 2. Its speed on landing, from $v=u+gt$ : $v = 0 + 9.8\times3 = 29.4\ \text{m/s}$ . So the dropped ball takes 3 s to fall and lands at 29.4 m/s.

Step 3. For the thrown stone, take upward as positive, so $u=19.6\ \text{m/s}$ , $g=-9.8\ \text{m/s}^2$ . At maximum height $v=0$ , so from $v^2=u^2+2gh$ :

0 = (19.6)^2 + 2(-9.8)h = 384.16 - 19.6h \implies h = \frac{384.16}{19.6}=19.6\ \text{m}

Step 4. Time to reach that height, from $v=u+gt$ with $v=0$ : $0 = 19.6+(-9.8)t \implies t = \frac{19.6}{9.8}=2\ \text{s}$ . By the up-down symmetry established above, the fall back to the hand takes the same 2 s, so the round trip is $T = 2\times2 = 4\ \text{s}$ .

The dropped ball falls 44.1 m in 3 s, landing at 29.4 m/s; the thrown stone rises to a maximum height of 19.6 m and returns to the hand after a total of 4 s.

Competitive stretch

If the same stone were thrown up from the edge of a 19.6 m tall tower instead of the ground, its height above the ground at the top of its flight would be $19.6+19.6 = 39.2\ \text{m}$ — the maximum rise above the launch point (19.6 m, unchanged, since it depends only on $u$ and $g$ ) simply adds to whatever height it was launched from.

Common traps

Dropping the negative sign on $g$ . Once upward is chosen as positive for a thrown object, $g$ must be substituted as $-9.8\ \text{m/s}^2$ in every equation used for that problem — forgetting this sign is the single most common error here.
Thinking the object is "not accelerating" at the top of its flight. Velocity is momentarily zero only at that one instant; $g$ never becomes zero, so the object is accelerating downward even while apparently paused.
Switching sign conventions mid-solution. Pick downward-positive or upward-positive once per problem, and keep every one of $u$ , $v$ , $g$ , $h$ consistent with that choice from start to finish.
Assuming time up always equals time down. That symmetry holds only when the object returns to the same height it was launched from and air resistance is ignored — both true for Class 9 problems, but worth stating rather than assuming blindly.

Ask the tutor: Want to try a mixed problem where a ball is thrown up from the edge of a cliff and you need both the maximum height above the cliff and the total time to hit the ground below — ask the AI tutor to walk through it with you.

Why does g have the same value whether an object is moving up, down, or momentarily at rest?

Gravity is a force that depends only on the masses involved and the distance between them (Earth's mass and the object's height above the surface) — not on how the object happens to be moving at that instant. Near Earth's surface this force always produces the same acceleration, about 9.8 m/s squared, directed toward Earth's centre, regardless of whether the object is rising, falling, or paused at its peak. Only the object's velocity changes with time; the acceleration causing that change stays constant throughout.

Why do we use h instead of s in these equations if the physics is identical?

It is purely a labelling convention to keep vertical problems visually distinct from horizontal ones — h reminds you the motion is vertical and the quantity is a height. The algebra, the derivation, and the underlying equations of motion are exactly the same three equations you already know for straight-line motion with constant acceleration; only the acceleration's value has been fixed to g and the symbol for distance renamed.

If I throw a ball up and it comes back to my hand, is its final velocity the same as its initial velocity?

Its speed (magnitude) is the same, by the up-down symmetry, but its velocity (which includes direction) is opposite in sign to the initial velocity — it left your hand moving upward and returns moving downward at the same speed, assuming air resistance is ignored, as is standard for Class 9 problems.

Motion of objects under gravity

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads