A Stone Thrown Up Takes Exactly as Long Coming Down
Learn to pick the right equation of motion for a numerical, find the distance covered in the nth second, solve free-fall and thrown-upward problems with the correct signs, and see why every body falls at the same rate.
Why does a stone take the same time going up as coming down?
Throw a stone straight up at . Gravity pulls it down at the whole time, so its upward speed drains away at every second:
Now it falls back from rest, with the same acting over the same height. So it builds up speed at exactly the rate it lost it, and takes exactly s to return.
It also arrives back at your hand at — the same speed you threw it with, in the opposite direction.
The reason is that the same acceleration acts over the same distance, once slowing the stone and once speeding it up. Nothing in the upward journey differs from the downward one except the sign of the velocity, which is why the two halves mirror each other exactly.
All of that comes out of three equations that the previous part of this chapter derived from a graph. This page applies them — to ordinary numericals, to the distance covered in one particular second, and to bodies falling and rising under gravity. It covers the third part of the ICSE Class 9 Physics chapter on motion in one dimension.
Now it falls back from rest, with the same acting over the same height. So it builds up speed at exactly the rate it lost it, and takes exactly s to return.
It also arrives back at your hand at — the same speed you threw it with, in the opposite direction.
The reason is that the same acceleration acts over the same distance, once slowing the stone and once speeding it up. Nothing in the upward journey differs from the downward one except the sign of the velocity, which is why the two halves mirror each other exactly.
All of that comes out of three equations that the previous part of this chapter derived from a graph. This page applies them — to ordinary numericals, to the distance covered in one particular second, and to bodies falling and rising under gravity. It covers the third part of the ICSE Class 9 Physics chapter on motion in one dimension.
Formula
How do you choose the right equation of motion for a problem?
List what you are given, note what is asked, and pick the equation that contains those quantities and nothing else. Each of the three leaves out exactly one variable.
So if the time is not given and not wanted, use the third. If the final velocity is irrelevant, use the second.
Worked example 1 — from rest. A car starts from rest with for s. Find its velocity and the distance covered.
Check with the third equation: , so . Consistent.
Worked example 2 — with an initial velocity. A cyclist at accelerates at for s.
Check: , so . Correct.
Worked example 3 — braking, with no time given. A car at brakes with a retardation of . How far does it go before stopping?
Time is neither given nor wanted, so use the third equation with :
And if the time is also wanted, the first equation gives , so s. Check with the second: m.
Worked example 4 — finding the acceleration from a distance. A train covers m while slowing from to .
a retardation of .
A retardation must be entered as a NEGATIVE acceleration. In worked example 3, putting would give m — a negative distance, which is the calculation telling you the sign was wrong. The formula does not know the car is braking; the sign is how you tell it.
Braking distance grows as the square of the speed. Since for a body braking to rest, doubling the speed quadruples the stopping distance. At the car above needed m; at it would need
Four times the distance for twice the speed, which is the physics behind every speed limit near a school.
So if the time is not given and not wanted, use the third. If the final velocity is irrelevant, use the second.
Worked example 1 — from rest. A car starts from rest with for s. Find its velocity and the distance covered.
Check with the third equation: , so . Consistent.
Worked example 2 — with an initial velocity. A cyclist at accelerates at for s.
Check: , so . Correct.
Worked example 3 — braking, with no time given. A car at brakes with a retardation of . How far does it go before stopping?
Time is neither given nor wanted, so use the third equation with :
And if the time is also wanted, the first equation gives , so s. Check with the second: m.
Worked example 4 — finding the acceleration from a distance. A train covers m while slowing from to .
a retardation of .
A retardation must be entered as a NEGATIVE acceleration. In worked example 3, putting would give m — a negative distance, which is the calculation telling you the sign was wrong. The formula does not know the car is braking; the sign is how you tell it.
Braking distance grows as the square of the speed. Since for a body braking to rest, doubling the speed quadruples the stopping distance. At the car above needed m; at it would need
Four times the distance for twice the speed, which is the physics behind every speed limit near a school.
How do you find the distance travelled in the nth second?
**Subtract the distance covered in seconds from the distance covered in seconds.** That difference is the ground covered during the th second alone.
Expanding and simplifying, the terms cancel and what remains is
Worked example 1 — from rest. A body starts from rest with . Find the distance in the th second.
Check the long way. In s: m. In s: m. The difference is m. Correct.
Worked example 2 — with an initial velocity. A body at with . Distance in the th second:
Check: in s, m; in s, m; difference m. Correct.
Worked example 3 — a falling body. A stone dropped from rest, with . Distance in the rd second:
Worked example 4 — the successive-seconds pattern. For the body in worked example 1, the distances in the first four seconds are
since . Dividing through by gives the ratio — the odd numbers, for any body starting from rest.
Check the total: m, and m for s. The pieces rebuild the whole.
"In the nth second" is not "in n seconds". The first is a distance covered during a one-second interval, and the second is the total from the start. For the body above, the distance in the th second is m while the distance in seconds is m. Those two numbers answer two different questions, and the single word the is what distinguishes them in the question.
The unit still comes out in metres, not metres per second. looks like a velocity because the formula is with a velocity as its first term. But is a pure count of seconds, so the expression is numerically a distance for a one-second interval. **Writing the answer as is a marked error** — it is m.
Expanding and simplifying, the terms cancel and what remains is
Worked example 1 — from rest. A body starts from rest with . Find the distance in the th second.
Check the long way. In s: m. In s: m. The difference is m. Correct.
Worked example 2 — with an initial velocity. A body at with . Distance in the th second:
Check: in s, m; in s, m; difference m. Correct.
Worked example 3 — a falling body. A stone dropped from rest, with . Distance in the rd second:
Worked example 4 — the successive-seconds pattern. For the body in worked example 1, the distances in the first four seconds are
since . Dividing through by gives the ratio — the odd numbers, for any body starting from rest.
Check the total: m, and m for s. The pieces rebuild the whole.
"In the nth second" is not "in n seconds". The first is a distance covered during a one-second interval, and the second is the total from the start. For the body above, the distance in the th second is m while the distance in seconds is m. Those two numbers answer two different questions, and the single word the is what distinguishes them in the question.
The unit still comes out in metres, not metres per second. looks like a velocity because the formula is with a velocity as its first term. But is a pure count of seconds, so the expression is numerically a distance for a one-second interval. **Writing the answer as is a marked error** — it is m.
How do you solve a free-fall or thrown-upward problem?
**Fix a positive direction, replace by with the matching sign, and use the same three equations. Nothing new is needed.
Taking downward as positive**, a falling body has . Taking upward as positive, a body thrown up has positive and .
Worked example 1 — dropped from rest. A stone is dropped and falls for s. Taking downward as positive with :
Worked example 2 — falling a known height. A stone is dropped from m. Find the time and the striking speed.
Check with the third equation: , and . Correct.
Worked example 3 — thrown vertically up. A ball is thrown up at . Take upward as positive, so .
Time to the top — at the highest point :
Maximum height:
Check with the second equation: m.
Total time of flight: s, by the symmetry this page opened with.
Velocity on return: — the same speed, downward.
Worked example 4 — a faster throw. Thrown up at :
and the total flight is s. Notice that m and are exactly the height and speed from worked example 2 — the stone dropped from m arrives at , and a ball thrown up at reaches m. The two problems are one problem run in opposite directions.
At the highest point the velocity is zero but the acceleration is NOT. The ball stops rising for an instant, and gravity is still pulling on it at the full — which is precisely why it does not stay there. **Writing at the top is the single commonest error in this topic, and it would predict a ball hanging in the air for ever.
Keep one sign convention for the whole problem.** Mixing a positive upward with a positive downward in the same line gives a wrong answer that looks arithmetically clean. State the choice in words before the first equation.
Taking downward as positive**, a falling body has . Taking upward as positive, a body thrown up has positive and .
Worked example 1 — dropped from rest. A stone is dropped and falls for s. Taking downward as positive with :
Worked example 2 — falling a known height. A stone is dropped from m. Find the time and the striking speed.
Check with the third equation: , and . Correct.
Worked example 3 — thrown vertically up. A ball is thrown up at . Take upward as positive, so .
Time to the top — at the highest point :
Maximum height:
Check with the second equation: m.
Total time of flight: s, by the symmetry this page opened with.
Velocity on return: — the same speed, downward.
Worked example 4 — a faster throw. Thrown up at :
and the total flight is s. Notice that m and are exactly the height and speed from worked example 2 — the stone dropped from m arrives at , and a ball thrown up at reaches m. The two problems are one problem run in opposite directions.
At the highest point the velocity is zero but the acceleration is NOT. The ball stops rising for an instant, and gravity is still pulling on it at the full — which is precisely why it does not stay there. **Writing at the top is the single commonest error in this topic, and it would predict a ball hanging in the air for ever.
Keep one sign convention for the whole problem.** Mixing a positive upward with a positive downward in the same line gives a wrong answer that looks arithmetically clean. State the choice in words before the first equation.
Why do a coin and a feather fall at the same rate in a vacuum?
**Because the acceleration of a falling body works out to with the mass cancelling out**, so nothing about the body itself enters the answer.
A body of mass in free fall has only its weight acting on it, and weight is . Its acceleration is the force divided by the mass:
The mass appears twice and cancels. A heavier body is pulled harder — which would speed it up — and is also harder to accelerate by exactly the same factor. The two effects are equal, so they leave nothing behind.
Worked comparison. A kg stone and a kg stone dropped together:
- kg: weight N, so
- kg: weight N, so
Ten times the pull, ten times the reluctance, identical acceleration. After s both are moving at and both have fallen m.
Why a feather loses in air. Air resistance is an extra upward force, and it depends on the body's shape and surface area rather than on its mass. A feather has a large area for very little weight, so the resistance is a big fraction of its weight and it drifts down slowly. A coin has a small area for its weight, so the resistance barely matters.
The everyday demonstration. Take two identical sheets of paper. Crumple one into a tight ball and drop both together. The ball lands well ahead, and their masses are identical — so mass cannot be the reason. Only the area changed. That experiment settles the question without any apparatus, and it is the answer to write when asked why a feather falls slowly.
Comparing the two halves of a vertical throw. For a ball thrown up and caught again at the same point:
- the time up equals the time down
- the speed at any given height is the same on both journeys, only reversed in direction
- the distance covered up equals the distance down, so the total distance is while the displacement is zero
- the acceleration is downward throughout — on the way up, at the top, and on the way down
Worked check on the speed symmetry. For the ball thrown at , find its speed at a height of m on the way up. Taking upward positive:
so upward. On the way down through the same height, the same equation with from the top and a fall of m gives
so downward. The same magnitude, the opposite direction — the symmetry confirmed by calculation rather than asserted.
The symmetry breaks the moment air resistance matters. With air resistance the ball comes down more slowly than it went up, and the downward journey takes longer, because resistance opposes the motion in both directions and so helps on the way up and hinders on the way down. Every result in this section assumes free fall, and saying so is part of a complete answer.
A body of mass in free fall has only its weight acting on it, and weight is . Its acceleration is the force divided by the mass:
The mass appears twice and cancels. A heavier body is pulled harder — which would speed it up — and is also harder to accelerate by exactly the same factor. The two effects are equal, so they leave nothing behind.
Worked comparison. A kg stone and a kg stone dropped together:
- kg: weight N, so
- kg: weight N, so
Ten times the pull, ten times the reluctance, identical acceleration. After s both are moving at and both have fallen m.
Why a feather loses in air. Air resistance is an extra upward force, and it depends on the body's shape and surface area rather than on its mass. A feather has a large area for very little weight, so the resistance is a big fraction of its weight and it drifts down slowly. A coin has a small area for its weight, so the resistance barely matters.
The everyday demonstration. Take two identical sheets of paper. Crumple one into a tight ball and drop both together. The ball lands well ahead, and their masses are identical — so mass cannot be the reason. Only the area changed. That experiment settles the question without any apparatus, and it is the answer to write when asked why a feather falls slowly.
Comparing the two halves of a vertical throw. For a ball thrown up and caught again at the same point:
- the time up equals the time down
- the speed at any given height is the same on both journeys, only reversed in direction
- the distance covered up equals the distance down, so the total distance is while the displacement is zero
- the acceleration is downward throughout — on the way up, at the top, and on the way down
Worked check on the speed symmetry. For the ball thrown at , find its speed at a height of m on the way up. Taking upward positive:
so upward. On the way down through the same height, the same equation with from the top and a fall of m gives
so downward. The same magnitude, the opposite direction — the symmetry confirmed by calculation rather than asserted.
The symmetry breaks the moment air resistance matters. With air resistance the ball comes down more slowly than it went up, and the downward journey takes longer, because resistance opposes the motion in both directions and so helps on the way up and hinders on the way down. Every result in this section assumes free fall, and saying so is part of a complete answer.
Exam tip
Exam tip: write u, v, a, S and t as a list before choosing an equation
List the five quantities and mark which are known. Then pick the equation missing the one you neither have nor want — when time is absent, when is.
State the sign convention in words first: taking downward as positive. Then keep it for the whole answer.
**A retardation goes in as a negative .** Using for braking gives a negative distance — that is the sign error announcing itself.
**Convert to with before substituting.
"In the nth second" means a one-second interval**, not the total: m in the th second against m in seconds.
**Use , and the answer is in metres**, never .
**At the highest point but still.** Never write at the top.
For a vertical throw, quote the symmetry: time up equals time down, and the return speed equals the throwing speed.
**Take unless the question says otherwise, and say which value you used.
Check with a second equation whenever there is time** — costs one line and catches most slips.
And give the unit and the direction on every answer: downward, m.
State the sign convention in words first: taking downward as positive. Then keep it for the whole answer.
**A retardation goes in as a negative .** Using for braking gives a negative distance — that is the sign error announcing itself.
**Convert to with before substituting.
"In the nth second" means a one-second interval**, not the total: m in the th second against m in seconds.
**Use , and the answer is in metres**, never .
**At the highest point but still.** Never write at the top.
For a vertical throw, quote the symmetry: time up equals time down, and the return speed equals the throwing speed.
**Take unless the question says otherwise, and say which value you used.
Check with a second equation whenever there is time** — costs one line and catches most slips.
And give the unit and the direction on every answer: downward, m.
Did you know
Why doubling your speed more than doubles the danger
A car braking to rest from a speed with a retardation covers
and the is the whole story. The stopping distance does not go up in step with the speed; it goes up with the square of it.
Take a car that can brake at :
- at (about ): m
- at (about ): m
- at (about ): m
Doubling the speed quadrupled the distance. Tripling it multiplied the distance by nine.
So a driver who speeds up from to has not made the road twice as risky — the braking distance went from about the length of three cars to about the length of twelve.
And the real stopping distance is worse than this, because the car also travels during the driver's reaction time before the brakes engage. That part does grow in step with the speed, so the two effects add: a linear reaction distance plus a squared braking distance.
The same shows up in the height a thrown ball reaches. Throwing at gives m; throwing at twice that, , gives
Four times the height for twice the throw. It is the same equation doing the same thing — makes distance depend on the square of the speed, whether the body is a braking car or a rising stone.
and the is the whole story. The stopping distance does not go up in step with the speed; it goes up with the square of it.
Take a car that can brake at :
- at (about ): m
- at (about ): m
- at (about ): m
Doubling the speed quadrupled the distance. Tripling it multiplied the distance by nine.
So a driver who speeds up from to has not made the road twice as risky — the braking distance went from about the length of three cars to about the length of twelve.
And the real stopping distance is worse than this, because the car also travels during the driver's reaction time before the brakes engage. That part does grow in step with the speed, so the two effects add: a linear reaction distance plus a squared braking distance.
The same shows up in the height a thrown ball reaches. Throwing at gives m; throwing at twice that, , gives
Four times the height for twice the throw. It is the same equation doing the same thing — makes distance depend on the square of the speed, whether the body is a braking car or a rising stone.
Exam relevance
How is uniformly accelerated motion tested in JEE Main and NEET?
Because the three equations are the most-used formulas in all of mechanics, and the free-fall case is the standard setting for problems in several later chapters.
This is the foundation for Class 11 Physics Motion in a Straight Line and Motion in a Plane, examined in JEE Main and NEET. The three equations are carried forward unchanged and then extended in two directions: derived by integration rather than from a graph, and applied in two dimensions as vector equations
Projectile motion is the direct extension of the vertical-throw work on this page. A projectile is a body with the horizontal motion of a uniform velocity and the vertical motion of exactly the throw analysed here, and the symmetry results — time up equals time down, the landing speed equals the launch speed — reappear as the properties of a projectile's path. Numericals on maximum height and time of flight are recurring JEE Main and NEET material, and their vertical half is this page.
**The dependence becomes the work-energy theorem.** Rearranging and multiplying by gives
which is the statement that work done equals the change in kinetic energy, met in Class 11 Work, Energy and Power. So the braking-distance result of the previous section is a kinetic-energy result in disguise, and that is why energy and stopping distance both go as the square of the speed.
The mass cancellation is revisited in Class 11 Gravitation, where the equality of gravitational pull and inertia is stated explicitly, and it is the reason an astronaut in orbit is weightless — a recurring assertion-reason topic in both exams.
**The th-second formula** appears as a shortcut, and the odd-number ratio turns some multiple-choice questions into a glance. JEE Main questions sometimes give distances in successive intervals and ask for the acceleration, which is this formula read backwards.
What the questions look like. For board work, expect numericals on all three equations, distance in the nth second, a stone dropped from a height, a ball thrown vertically up asking for maximum height and time of flight, and why all bodies fall alike as a reasoning question with the crumpled-paper example. For JEE Main and NEET, expect projectiles, motion under gravity combined with relative velocity, graph-based questions, and energy versions of the same problems.
How board and competitive emphasis differ. A board paper rewards the listed quantities, the stated sign convention and the substitution written out. A competitive paper assumes all three equations and tests which one reaches the answer in fewest steps — often the third, because it avoids finding the time.
The single trap that costs the most marks. Setting the acceleration to zero at the top of a vertical throw. The velocity is zero there for an instant, and the acceleration is the full downward throughout — which is exactly why the ball does not hover. **The defence is to write once at the top of the answer and never touch it again**, changing only and between parts.
This is the foundation for Class 11 Physics Motion in a Straight Line and Motion in a Plane, examined in JEE Main and NEET. The three equations are carried forward unchanged and then extended in two directions: derived by integration rather than from a graph, and applied in two dimensions as vector equations
Projectile motion is the direct extension of the vertical-throw work on this page. A projectile is a body with the horizontal motion of a uniform velocity and the vertical motion of exactly the throw analysed here, and the symmetry results — time up equals time down, the landing speed equals the launch speed — reappear as the properties of a projectile's path. Numericals on maximum height and time of flight are recurring JEE Main and NEET material, and their vertical half is this page.
**The dependence becomes the work-energy theorem.** Rearranging and multiplying by gives
which is the statement that work done equals the change in kinetic energy, met in Class 11 Work, Energy and Power. So the braking-distance result of the previous section is a kinetic-energy result in disguise, and that is why energy and stopping distance both go as the square of the speed.
The mass cancellation is revisited in Class 11 Gravitation, where the equality of gravitational pull and inertia is stated explicitly, and it is the reason an astronaut in orbit is weightless — a recurring assertion-reason topic in both exams.
**The th-second formula** appears as a shortcut, and the odd-number ratio turns some multiple-choice questions into a glance. JEE Main questions sometimes give distances in successive intervals and ask for the acceleration, which is this formula read backwards.
What the questions look like. For board work, expect numericals on all three equations, distance in the nth second, a stone dropped from a height, a ball thrown vertically up asking for maximum height and time of flight, and why all bodies fall alike as a reasoning question with the crumpled-paper example. For JEE Main and NEET, expect projectiles, motion under gravity combined with relative velocity, graph-based questions, and energy versions of the same problems.
How board and competitive emphasis differ. A board paper rewards the listed quantities, the stated sign convention and the substitution written out. A competitive paper assumes all three equations and tests which one reaches the answer in fewest steps — often the third, because it avoids finding the time.
The single trap that costs the most marks. Setting the acceleration to zero at the top of a vertical throw. The velocity is zero there for an instant, and the acceleration is the full downward throughout — which is exactly why the ball does not hover. **The defence is to write once at the top of the answer and never touch it again**, changing only and between parts.
Key takeaways
Equations of motion, the nth second and free fall: quick revision
- The three equations: (no ), (no ), (no ). Pick the one missing the variable you neither have nor want.
- Rest with for s: , m, confirmed by .
- at for s: , m, confirmed by .
- Braking from at : m and s.
- Slowing from to over m: , a retardation of .
- **A retardation goes in as a negative — a positive value returns a negative distance.
- Braking distance is , so doubling the speed quadruples it**: m at becomes m at .
- Distance in the nth second: , obtained by subtracting the -second distance from the -second one.
- From rest at , the th second gives m, checked as .
- At , , the th second gives m, checked as .
- A dropped stone covers m in the rd second.
- Successive seconds from rest go as **** — m, totalling m.
- "In the nth second" is a one-second interval, in metres — m, never , and not the m of five seconds.
- Free fall: dropped for s gives and m.
- Dropped from m: s and , confirmed by .
- **Thrown up at **: s, m, flight s, return velocity .
- **Thrown up at **: s, m, flight s — the drop problem run backwards.
- **At the top but still.** Never set there.
- **All bodies fall alike because ** — the mass cancels. A kg and a kg stone both accelerate at .
- Air resistance depends on shape and area, not mass — two identical sheets of paper, one crumpled, prove it.
- Up-and-down symmetry: equal times, equal speeds at each height, total distance with zero displacement, and downward throughout.
- At m the ball moves at on both journeys, since either way.
- The symmetry assumes free fall — with air resistance the descent takes longer.
Drop a flat sheet of paper and a crumpled one from the same height and time them, then explain the difference using only their shapes — their masses are identical.
- Rest with for s: , m, confirmed by .
- at for s: , m, confirmed by .
- Braking from at : m and s.
- Slowing from to over m: , a retardation of .
- **A retardation goes in as a negative — a positive value returns a negative distance.
- Braking distance is , so doubling the speed quadruples it**: m at becomes m at .
- Distance in the nth second: , obtained by subtracting the -second distance from the -second one.
- From rest at , the th second gives m, checked as .
- At , , the th second gives m, checked as .
- A dropped stone covers m in the rd second.
- Successive seconds from rest go as **** — m, totalling m.
- "In the nth second" is a one-second interval, in metres — m, never , and not the m of five seconds.
- Free fall: dropped for s gives and m.
- Dropped from m: s and , confirmed by .
- **Thrown up at **: s, m, flight s, return velocity .
- **Thrown up at **: s, m, flight s — the drop problem run backwards.
- **At the top but still.** Never set there.
- **All bodies fall alike because ** — the mass cancels. A kg and a kg stone both accelerate at .
- Air resistance depends on shape and area, not mass — two identical sheets of paper, one crumpled, prove it.
- Up-and-down symmetry: equal times, equal speeds at each height, total distance with zero displacement, and downward throughout.
- At m the ball moves at on both journeys, since either way.
- The symmetry assumes free fall — with air resistance the descent takes longer.
Drop a flat sheet of paper and a crumpled one from the same height and time them, then explain the difference using only their shapes — their masses are identical.