PhysicsCBSE Class 11 · Motion in a Straight Line

Free Fall, Vertical Throws and Stopping Distance Class 11

Updated 17 October 2026 · Fact-checked

Free fall is motion under gravity alone with constant acceleration g (taken as 10 m/s² here, directed downwards). A body thrown up at speed u reaches a height u²/2g after u/g seconds and returns after 2u/g. Stopping distance of a vehicle is the reaction distance (speed times reaction time) plus the braking distance v²/2a.

Understand Free Fall, Vertical Throws and Stopping Distance

Near the Earth's surface, every body in the absence of air resistance has the same downward acceleration $g \approx 9.8 m/s^{2}$ ; in this page problems take $g = 10 m/s^{2}$ unless stated. Free fall therefore uses the same equations as any uniform motion, with $a = \pm g$ .

Choosing signs. Take upward as positive: $a = - g$ . For a body dropped from rest from height $H$ : $t = 2 H / g$ , final speed $v = 2 g H$ . The distances in successive seconds from rest are in the ratio $1 : 3 : 5 : \dots$ , and the distance in the $n$ th second is $\frac{g}{2} (2 n - 1)$ .

Vertical throw upwards with speed $u$ from the ground:

time to the top $t = u / g$ ; maximum height $H = u^{2} /2 g$ ;
time of flight $T = 2 u / g$ ; return speed equals $u$ (downward);
at height $h$ below the top the speed is the same going up and coming down, and the body passes each height twice (once up, once down) except at the top.

At the highest point the velocity is zero but the acceleration is still $g$ downward.

Dropping from a moving carrier. A stone released from a rising balloon starts with the balloon's upward velocity, not from rest; use $u = +$ speed of the balloon and $s = -$ height of release.

Stopping distance. A driver takes a reaction time $t_{r}$ to apply brakes, during which the car moves at constant speed $v$ : reaction distance $= v t_{r}$ . The braking distance with deceleration $a$ is $v^{2} /2 a$ . Total stopping distance $= v t_{r} + \frac{v ^{2}}{2 a}$ . It grows with the square of speed, which is why speed limits matter. Reaction time can be estimated by dropping a ruler: $t = 2 d / g$ where $d$ is the distance fallen before it is caught.

Free Fall, Vertical Throws and Stopping Distance: formulas and key facts

Fall from height H: $t = \frac{2 H}{g}, v = 2 g H$; From rest.
Maximum height: $H = \frac{u ^{2}}{2 g}$; Thrown up at u.
Time of flight: $T = \frac{2 u}{g}$; Ground to ground.
Distance in nth second (from rest): $s_{n} = \frac{g}{2} (2 n - 1)$; Ratio 1:3:5.
Stopping distance: $d = v t_{r} + \frac{v ^{2}}{2 a}$; Reaction plus braking.
Reaction time from a ruler drop: $t = \frac{2 d}{g}$; Distance d fallen.

How to solve Free Fall, Vertical Throws and Stopping Distance questions

Fix the origin and the positive direction before substituting; most mistakes are sign mistakes.

1Draw the path and mark the origin, usually the point of projection.
2Take upward positive, so a = −g, and substitute signed values for u and s.
3For two bodies, write each position y(t) from a common origin and equate for a meeting.
4Use v = 0 at the highest point; time to the top = u/g.
5For a quadratic in t, check both roots and reject the unphysical one.
6For stopping distance, convert km/h to m/s first (divide by 3.6).

Free Fall, Vertical Throws and Stopping Distance 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.

A phone dropped from a balcony

Suppose a phone slips from a balcony 20 m above the pavement and you take g = 10 m/s².

Find the fall time and the speed on hitting the ground.

Try it first, then show the solutionSolution

t = √(2 × 20/10) = 2 s.
v = gt = 20 m/s.

Answer: 2 s; 20 m/s.

Memory hook: Fall time grows as the square root of height.

Braking before a zebra crossing

Suppose a scooter at 36 km/h, reaction time 0.6 s, deceleration 4 m/s².

Find the stopping distance.

Try it first, then show the solutionSolution

36 km/h = 10 m/s.
Reaction distance = 10 × 0.6 = 6 m; braking distance = 100/(2 × 4) = 12.5 m.
Total = 18.5 m.

Answer: 18.5 m.

Memory hook: Double the speed and the braking part grows four times.

Tricks to remember Free Fall, Vertical Throws and Stopping Distance

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

Up and down are mirror images
Time up = time down, speed back at the start = launch speed.
Use it when: Time of flight problems.
Top: v zero, a = g
Zero velocity does not mean zero acceleration.
Use it when: Conceptual traps.
Last-second trick
Last-second distance from rest: g(2n − 1)/2, so match it with n.
Use it when: Distance in a given second.
Square of speed
Braking distance ∝ v².
Use it when: Road-safety reasoning.

Free Fall, Vertical Throws and Stopping Distance: solved examples

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

Example 1: Dropped from a tower · 2 marks · Very short answerBoard level

A coconut falls from rest from a branch 11.25 m above the ground (g = 10 m/s²). Find the time to reach the ground and the speed on impact.

Try it first, then show the solutionSolution

t = √(2H/g) = √(2 × 11.25/10) = √2.25 = 1.5 s. (1)
v = gt = 10 × 1.5 = 15 m/s. (1)

Answer: 1.5 s; 15 m/s.

Example 2: Thrown upward · 3 marks · Short answerHarder board level

A ball is thrown vertically up at 20 m/s (g = 10 m/s²). At what times is it 15 m above the point of projection, and what is its velocity then?

Try it first, then show the solutionSolution

15 = 20t − 5t² gives t² − 4t + 3 = 0, so t = 1 s and 3 s. (1)
v = 20 − 10t: at t = 1 s, v = +10 m/s (upward). (1)
At t = 3 s, v = −10 m/s (downward); same speed at the same height. (1)

Answer: t = 1 s (10 m/s up) and 3 s (10 m/s down).

Example 3: Stopping distance of a car · 3 marks · Short answerHarder board level

A car travels at 54 km/h. The driver's reaction time is 0.8 s. Find the total stopping distance (a) on a dry road with deceleration 3 m/s², (b) on a wet road with deceleration 1.5 m/s².

Try it first, then show the solutionSolution

v = 54/3.6 = 15 m/s. Reaction distance = 15 × 0.8 = 12 m. (1)
(a) Braking = 225/(2 × 3) = 37.5 m, total 49.5 m. (1)
(b) Braking = 225/(2 × 1.5) = 75 m, total 87 m. (1)

Answer: (a) 49.5 m; (b) 87 m.

Example 4: Separation of two stones · 3 marks · Short answerHarder board level

A stone is dropped from a bridge. One second later a second stone is dropped from the same point (g = 10 m/s²). Find the separation between them 3 s after the first is released, and show that it keeps growing.

Try it first, then show the solutionSolution

First stone: ½ × 10 × 9 = 45 m. Second (2 s of fall): ½ × 10 × 4 = 20 m. (1)
Separation = 45 − 20 = 25 m. (1)
In general 5t² − 5(t − 1)² = 5(2t − 1) m, which increases with t. (1)

Answer: 25 m; the gap is 5(2t − 1) m.

Example 5: Two bodies meet · 4 marks · Long answerHOTS / competency

A ball A is dropped from the top of a 120 m tower at the same instant a ball B is thrown vertically upward from the foot of the tower at 40 m/s (g = 10 m/s²). Find when and where they meet, and the speed of each at that instant.

Try it first, then show the solutionSolution

Take upward positive: yA = 120 − 5t², yB = 40t − 5t². (1)
Equal: 120 = 40t, so t = 3 s. (1)
Height = 40 × 3 − 5 × 9 = 75 m above the ground. (1)
vA = −10 × 3 = −30 m/s (30 m/s down); vB = 40 − 30 = +10 m/s (10 m/s up). (1)

Answer: t = 3 s at 75 m; A at 30 m/s down, B at 10 m/s up.

Where students lose marks in Free Fall, Vertical Throws and Stopping Distance

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

Mistake: Taking the initial speed of a stone dropped from a rising balloon as zero.
Ignoring the carrier's velocity.
Do this: u equals the balloon speed, upward.
Usually costs: The setup and answer marks.
Mistake: Mixing g = +10 and −10 in one line.
No sign convention.
Do this: State "upward positive, a = −g" first.
Usually costs: 1 to 2 marks.
Mistake: Forgetting to convert km/h to m/s.
Using 72 directly.
Do this: Divide by 3.6 before substituting.
Usually costs: Accuracy mark.
Mistake: Writing the acceleration as zero at the top.
Confusing v = 0 with a = 0.
Do this: At the top a is still g downward.
Usually costs: Reasoning mark.

Exam tips

Write "take upward as positive" as your first line.
Reject the negative or smaller time root that does not suit the question.
Give the speed at hitting the ground as a magnitude and state the direction.
State g = 10 or 9.8 m/s² explicitly; keep the same value throughout.
For stopping distance always show reaction distance and braking distance separately.

Practice questions on Free Fall, Vertical Throws and Stopping Distance

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
A ball thrown vertically up returns to the hand after 6 s (g = 10 m/s²). Its maximum height is
(a) 15 m (b) 30 m (c) 45 m (d) 90 m
Hint
Time up = 3 s.
Answer
(c) 45 m: u = 30 m/s, H = 900/20 = 45 m.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): At the highest point of its flight, the acceleration of a ball thrown upward is zero.
Reason (R): Its velocity is zero at that point.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Gravity keeps acting.
Answer
(d). The velocity is zero but the acceleration remains g downward.
Q3 · 2 marks · Very short answerBoard level
Two stones of mass 1 kg and 5 kg are dropped together from the same height in vacuum. Compare their times of fall and their speeds on landing, and give the reason.
Hint
Mass does not appear in t = √(2H/g).
Answer
Both take the same time and land with the same speed (√(2gH)), because the acceleration g does not depend on mass.
Q4 · 3 marks · Short answerHarder board level
A student catches a falling ruler after it has fallen 20 cm. Find her reaction time (g = 10 m/s²). What would the fall be for a reaction time of 0.3 s?
Hint
d = ½gt².
Answer
t = √(2 × 0.2/10) = 0.2 s. For 0.3 s: d = ½ × 10 × 0.09 = 0.45 m = 45 cm.
Q5 · 4 marks · Long answerHOTS / competency
A balloon rises at a constant 10 m/s. When it is 75 m above the ground a stone is dropped from it (g = 10 m/s²). Find (a) the time for the stone to reach the ground, (b) its speed on landing, (c) the greatest height above the ground it reaches.
Hint
The stone starts with +10 m/s.
Answer
(a) −75 = 10t − 5t² gives t² − 2t − 15 = 0, t = 5 s. (b) v = 10 − 50 = −40, i.e. 40 m/s downward. (c) Extra rise = 100/20 = 5 m, so 80 m.

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.Time to reach the top?Show answer
u/g.
2.Maximum height?Show answer
u²/2g.
3.Acceleration at the top?Show answer
g downward.
4.Total stopping distance?Show answer
vt_r + v²/2a.
5.72 km/h in m/s?Show answer
20 m/s.
6.Distance in nth second from rest?Show answer
g(2n − 1)/2.

Free Fall, Vertical Throws and Stopping Distance: frequently asked questions

Do heavier bodies fall faster?

In the absence of air resistance, no: all bodies have the same acceleration g.

Why does the stone from a balloon first go up?

It leaves with the balloon's upward velocity, so it continues upward until gravity stops it.

How does stopping distance change if the speed is doubled?

The braking distance becomes four times and the reaction distance doubles.

Which equation do I use for the time of flight?

Set s = 0 in s = ut − ½gt² to get T = 2u/g for ground-to-ground flight.

Why take g = 10?

It makes the arithmetic clean; 9.8 m/s² is the more accurate value and answers change slightly.