PhysicsICSE Class 9 · Motion in One Dimension

Equations of Motion and Free Fall Under Gravity: Solved Numericals (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

For uniform acceleration along a straight line use v = u + at, S = ut + ½at², v² = u² + 2aS and S = ½(u + v)t. Pick the equation that leaves out the quantity you neither know nor need. For free fall take a = +g downward (g = 10 m s⁻² here) and keep one sign convention throughout.

Speed grows by 10 m/s every second, so the distances fallen in successive seconds, 5, 15 and 25 m, are in the ratio 1 : 3 : 5.

Illustration drawn for QuestionBuddy.

Understand Equations of Motion and Free Fall Under Gravity

Symbols. $u$ = initial velocity, $v$ = final velocity, $a$ = uniform acceleration, $t$ = time, $S$ = displacement. They hold only for uniform acceleration in a straight line (non-uniform acceleration is outside the Class IX syllabus).

Equation	Missing quantity
$v = u + a t$	$S$
$S = u t + \frac{1}{2} a t^{2}$	$v$
$v^{2} = u^{2} + 2 a S$	$t$
$S = \frac{u + v}{2} t$	$a$

Choosing the equation. List what you are given and what is asked. Find the one quantity of the five ( $u, v, a, t, S$ ) that is neither given nor asked: that is the missing quantity, and the equation in the table without it is the one to use.

Sign convention. Pick a positive direction and keep to it. A body slowing down has negative $a$ (retardation) if $u$ is positive. Convert km h⁻¹ to m s⁻¹ by multiplying by $\frac{5}{18}$ .

**Distance in the $n$ th second.** Distance in the $n$ th second is the difference between the distances in $n$ seconds and in $(n - 1)$ seconds. Using $S = u t + \frac{1}{2} a t^{2}$ for both gives

$S_{n} = u + \frac{a}{2} (2 n - 1) .$

This is a result obtained by subtraction; learn the method (subtract) rather than only the formula.

Free fall. A body released and moved only by gravity has $u = 0$ and $a = g \approx 10$ m s⁻² downward (9.8 m s⁻² more exactly). Then $v = g t$ , $h = \frac{1}{2} g t^{2}$ , $v^{2} = 2 g h$ . The distances fallen in successive equal seconds from rest are in the ratio $1 : 3 : 5 : \dots$ .

Thrown up. With upward positive, $a = - g$ . At the highest point $v = 0$ but $a$ is still $g$ downward. Time up equals time down (to the same level), and the velocity on return to the starting level has the same size as the velocity of projection but is directed downward:

$H = \frac{u ^{2}}{2 g}, t_{u p} = \frac{u}{g}, t_{t o t a l} = \frac{2 u}{g} .$

Two bodies. For problems with two moving bodies, write the position of each against the same clock and use a condition (meeting: positions equal, or distances add to the total gap).

Equations of Motion and Free Fall Under Gravity: formulas and key facts

Equations of motion: $v = u + a t; S = u t + \frac{1}{2} a t^{2}; v^{2} = u^{2} + 2 a S; S = \frac{u + v}{2} t$; Uniform acceleration only.
Distance in the nth second: $S_{n} = u + \frac{a}{2} (2 n - 1)$; Obtained as S(n) − S(n−1).
Free fall from rest: $v = g t; h = \frac{1}{2} g t^{2}; v^{2} = 2 g h$; g = 10 m s⁻² (or 9.8).
Thrown vertically up: $H = \frac{u ^{2}}{2 g}; t_{u p} = \frac{u}{g}; t_{t o t a l} = \frac{2 u}{g}$; Same level to same level.

How to solve Equations of Motion and Free Fall Under Gravity questions

Treat every motion problem as a list of five symbols.

1Convert all quantities to SI (m, s, m s⁻¹).
2Write u, v, a, t, S with signs; mark the unknown.
3Find the missing quantity and choose the matching equation.
4Substitute with units, solve, then check the sign and size make sense.
5For two-body problems write both positions against the same time.

Equations of Motion and Free Fall Under Gravity in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Dropping a coin into a well

Suppose you drop a coin into a well and, from the moment you release it, you hear the splash after 2 s. Ignore the travel time of sound.

How deep is the water surface below you, and how fast is the coin going at the splash? (g = 10 m s⁻²)

Try it first, then show the solutionSolution

$h = \frac{1}{2} g t^{2} = 5 \times 4 = 20$ m.
$v = g t = 10 \times 2 = 20$ m s⁻¹.

Answer: 20 m deep; 20 m s⁻¹ downward at the splash.

Memory hook: Fall time squared times five (for g = 10).

Tricks to remember Equations of Motion and Free Fall Under Gravity

Short memory aids. Say each one aloud once, then use it on a question the same day.

Five symbols, one missing
u v a t S: name the one you do not have or want; that picks the equation.
Use it when: Choosing equations.
Up and down are mirror images
Time up = time down; speed back at the start level = speed of throw.
Use it when: Thrown-up bodies.
Odd-number rule
Falling from rest: distances in successive seconds are 1 : 3 : 5 : 7.
Use it when: Quick checks.

Equations of Motion and Free Fall Under Gravity: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Distance fallen in 3 s · 1 mark · MCQBoard level

A stone is dropped from rest. The distance it falls in 3 s ( $g = 10$ m s⁻²) is:
(a) 15 m
(b) 30 m
(c) 45 m
(d) 90 m

Try it first, then show the solutionSolution

$h = \frac{1}{2} g t^{2} = 5 \times 9 = 45$ m.

Answer: (c) 45 m

Example 2: A train braking · 2 marks · Short answerBoard level

A train moving at 108 km h⁻¹ is brought to rest uniformly in 12 s. Find the retardation and the distance covered.

Try it first, then show the solutionSolution

$u = 108 \times \frac{5}{18} = 30$ m s⁻¹; $v = 0$ ; $t = 12$ s. (½)
$a = \frac{0 - 30}{12} = - 2.5$ m s⁻². (½)
$S = \frac{30 + 0}{2} \times 12 = 180$ m. (1)

Answer: Retardation 2.5 m s⁻²; distance 180 m

Example 3: Stone thrown down from a tower · 3 marks · Short answerHarder board level

A stone is thrown vertically downward with a speed of 5 m s⁻¹ from a tower 60 m high. Find the time to reach the ground and its speed on landing ( $g = 10$ m s⁻²).

Try it first, then show the solutionSolution

$60 = 5 t + 5 t^{2}$ , i.e. $t^{2} + t - 12 = 0$ . (1)
$(t + 4) (t - 3) = 0$ , so $t = 3$ s (reject $- 4$ ). (1)
$v = u + g t = 5 + 30 = 35$ m s⁻¹. (1)

Answer: 3 s; 35 m s⁻¹

Example 4: Fourth-second distance · 3 marks · Short answerHarder board level

A body starts from rest and moves with uniform acceleration. It covers 14 m in the 4th second. Find its acceleration and the distance covered in 6 s.

Try it first, then show the solutionSolution

$S_{4} = \frac{a}{2} (2 \times 4 - 1) = 3.5 a = 14$ , so $a = 4$ m s⁻². (1½)
Check: $\frac{1}{2} (4) (16) - \frac{1}{2} (4) (9) = 32 - 18 = 14$ m. (½)
$S_{6} = \frac{1}{2} \times 4 \times 36 = 72$ m. (1)

Answer: a = 4 m s⁻²; 72 m

Example 5: Two stones, one meeting · 4 marks · Case-basedHOTS / competency

A stone A is dropped from the top of an 80 m tower. At the same instant a stone B is thrown vertically upward from the foot of the tower with a speed of 40 m s⁻¹ ( $g = 10$ m s⁻²).
(a) After what time do they meet?
(b) At what height above the ground?
(c) Find the velocity of each at that instant.
(d) Show that if B were thrown up with speed $u$ they would meet after $\frac{80}{u}$ seconds.

Try it first, then show the solutionSolution

(a) Distances add to 80 m: $\frac{1}{2} g t^{2} + (40 t - \frac{1}{2} g t^{2}) = 40 t = 80$ , so $t = 2$ s. (1)
(b) $h = 40 (2) - 5 (4) = 60$ m (A has fallen 20 m, 80 − 20 = 60). (1)
(c) A: $10 \times 2 = 20$ m s⁻¹ down; B: $40 - 20 = 20$ m s⁻¹ up. (1)
(d) The meeting time is $\frac{80}{u _{B}}$ because the gravitational terms cancel; a larger $u_{B}$ gives an earlier meeting at a different height. (1)

Answer: (a) 2 s (b) 60 m (c) 20 m s⁻¹ down and 20 m s⁻¹ up (d) t = 80/u_B

Where students lose marks in Equations of Motion and Free Fall Under Gravity

The slips examiners see most often, and what to write instead.

Mistake: Retardation given a positive sign in the equation
With u positive, slowing down means a is negative.
Wrong working
v² = u² + 2aS with a = +5 gives a bigger v.
Right working
a = −5 m s⁻²; 0 = u² − 10S.
Do this: Write each quantity with its sign.
Usually costs: The whole numerical.
Mistake: Speed in km h⁻¹ substituted with time in seconds
Units must be SI.
Wrong working
a = (0 − 108)/12 = −9
Right working
108 km h⁻¹ = 30 m s⁻¹; a = −2.5 m s⁻².
Do this: Multiply km h⁻¹ by 5/18.
Usually costs: Answer and one method mark.
Mistake: Quadratic root not rejected
Time cannot be negative.
Wrong working
t = 3 s or −4 s (both reported).
Right working
t = 3 s only.
Do this: State why the other root is rejected.
Usually costs: 1 mark.

Exam tips

Write the five symbols and mark the unknown before choosing an equation.
Show the unit conversion line explicitly.
Say "taking upward as positive, a = −g" at the start of a thrown-up question.
For two-body problems, state the condition (equal positions or distances adding to the gap).

Practice questions on Equations of Motion and Free Fall Under Gravity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
At the highest point of a ball thrown vertically upward, its velocity and acceleration are respectively:
(a) zero and zero
(b) zero and g downward
(c) g and zero
(d) maximum and g upward
Hint
Gravity still acts.
Answer
(b) zero and g downward
Q2 · 1 mark · Very short answerBoard level
A ball dropped from rest reaches a speed of 25 m s⁻¹ just before landing. Find the time of fall ( $g = 10$ m s⁻²).
Hint
v = gt.
Answer
$t = \frac{25}{10} = 2.5$ s.
Q3 · 2 marks · Short answerHarder board level
A ball thrown vertically upward returns to the thrower’s hand after 5 s. Find the speed of projection and the maximum height ( $g = 10$ m s⁻²).
Hint
Up-time is half of the total.
Answer
$t_{u p} = 2.5$ s; $u = 10 \times 2.5 = 25$ m s⁻¹; $H = \frac{2 5 ^{2}}{2 \times 10} = 31.25$ m.
Q4 · 3 marks · Short answerHarder board level
A cyclist at 36 km h⁻¹ accelerates uniformly at 0.5 m s⁻² for 20 s. Find the final speed, the distance covered and the average speed.
Hint
Convert 36 km h⁻¹ first.
Answer
$u = 10$ m s⁻¹; $v = 10 + 0.5 \times 20 = 20$ m s⁻¹; $S = 10 \times 20 + \frac{1}{2} (0.5) (400) = 300$ m; average speed $= \frac{300}{20} = 15$ m s⁻¹.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): A body can have zero velocity and yet non-zero acceleration.
Reason (R): Acceleration is the rate of change of velocity.
(a) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(b) Both are true, but (R) is not the correct explanation.
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Think of the top of a throw.
Answer
(a). At the top of a vertical throw v = 0, but velocity is still changing, so a = g.
Q6 · 3 marks · Short answerHOTS / competency
Ball P is dropped from a height; exactly 1 s later ball Q is dropped from the same point ( $g = 10$ m s⁻²). Find the separation of the balls when P has fallen 45 m, and show that the separation keeps increasing.
Hint
Write both distances against the same clock.
Answer
P: $45 = 5 t^{2}$ , $t = 3$ s; Q has fallen for 2 s: $5 \times 4 = 20$ m; separation $= 25$ m. In general separation $= 5 t^{2} - 5 (t - 1)^{2} = 10 t - 5$ , which grows with $t$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Equation with no time?Show answer
v² = u² + 2aS.
2.Time to reach the top of a throw at speed u?Show answer
u/g.
3.Distance in the nth second?Show answer
S(n) − S(n−1) = u + a(2n − 1)/2.
4.Acceleration at the top of a throw?Show answer
g downward (not zero).
5.Convert 72 km h⁻¹ to m s⁻¹.Show answer
20 m s⁻¹.

Equations of Motion and Free Fall Under Gravity: frequently asked questions

Which equation of motion should I use?

Find the quantity among u, v, a, t, S that is neither given nor asked, and use the equation that does not contain it.

What is the value of g to use in ICSE Class 9?

Use the value stated in the question, usually 10 m s⁻² (or 9.8 m s⁻² for more precise work).

Is acceleration zero at the top of a vertical throw?

No. The velocity is zero for an instant, but gravity still gives an acceleration g downward.