Solve free-fall problems involving gravity and acceleration

Ontario Grade 11 Physics

Forces

Gravity, acceleration, and choosing a consistent direction

A ball released from a hand speeds up as it falls. A ball tossed upward slows down, stops briefly, and then falls. In both cases, gravity causes a downward acceleration. Free-fall calculations become clearer when you first choose a positive direction and then use it consistently. This lesson uses the Grade 11 constant-acceleration relationships to solve problems involving gravity.

What you will learn

Describe free fall using a model in which gravity is the only force considered.
Choose and use a positive direction to represent velocity and acceleration.
Select a constant-acceleration relationship to solve for an unknown in a free-fall problem.
Check the units, direction, significant figures, and reasonableness of a result.

1. Prerequisite bridge: quantities and signs

A scalar has size only. Time is a scalar: a time interval could be

2.0\ \mathrm{s}

. A vector has size and direction. Velocity, displacement, and acceleration are vectors. For example, a velocity can be

3.0\ \mathrm{m/s}

downward.

A coordinate direction is the direction chosen to count as positive. For vertical motion, choose either upward or downward as positive. The opposite direction is negative. This choice is not a claim about which direction is naturally positive; it is a bookkeeping tool.

Acceleration describes how velocity changes over time. Near Earth's surface, the acceleration due to gravity has a magnitude of about

9.81\ \mathrm{m/s^2}

and points downward. The symbol

g

usually means this positive magnitude. The signed acceleration

a

depends on your choice of positive direction.

g=9.81\ \mathrm{m/s^2}

State the positive direction before assigning signs.
Use seconds for time, metres for displacement, metres per second for velocity, and metres per second squared for acceleration.
A negative answer means the quantity points opposite to the chosen positive direction.

2. The free-fall model and its diagram

The physical system in a free-fall problem is the object whose motion is being described. In the free-fall model, gravity is the only force considered to affect the object's motion. This model ignores the effect of air resistance. Use it when a problem says to assume free fall or gives no reason to account for air resistance.

For a chosen vertical coordinate, the acceleration stays constant during the motion. If upward is positive, the acceleration is

-9.81\ \mathrm{m/s^2}

. If downward is positive, it is

+9.81\ \mathrm{m/s^2}

. The object can move upward while accelerating downward; velocity and acceleration do not have to point in the same direction.

Before calculating, identify the known quantities and the unknown. Record the initial velocity

v_i

, final velocity

v_f

, elapsed time

t

, displacement

\Delta d

, and acceleration

a

. Initial means at the start of the interval; final means at the end. Displacement is the change in position, with a sign set by the coordinate direction.

A simple vector diagram for upward-positive coordinates is: upward is positive; the velocity of a rising object points upward; the acceleration due to gravity points downward. This means the rising object's velocity is positive while its acceleration is negative. At its highest point, its velocity is momentarily zero, but its acceleration is still downward.

a=-g\ \text{(upward positive)},\qquad a=+g\ \text{(downward positive)}

Free fall uses a constant downward acceleration in this model.
The sign of acceleration changes with the chosen positive direction, not with whether the object is moving up or down.
At the highest point of an upward toss, velocity is zero for an instant; gravity's acceleration is not zero.

3. Choose a relationship and solve

Use a constant-acceleration relationship that contains the unknown and the quantities you know. These relationships apply to the free-fall model because its acceleration is constant. Keep the signs and units in every substitution. The symbol

\Delta d

means final position minus initial position, measured along the chosen vertical coordinate.

If time is known, the velocity relationship can connect initial velocity, final velocity, acceleration, and time. The displacement relationship can connect displacement to initial velocity, acceleration, and time. If time is not known or needed, the relationship involving the two velocities, acceleration, and displacement can be useful.

Rearrange the selected relationship with ordinary algebra before inserting values. Use SI units. Round the final result to a sensible number of significant figures based on the given values. Include the direction in words, especially when the signed result is negative.

v_f=v_i+at,\qquad \Delta d=v_i t+\frac{1}{2}at^2,\qquad v_f^2=v_i^2+2a\Delta d

Match the equation to the known values and the requested unknown.
Do not use a positive value for gravity automatically; determine the signed acceleration from your coordinate choice.
Check that the answer has the expected units and points in a physically sensible direction.

4. Check the result

A unit check can reveal a setup error. For example, in the velocity relationship, acceleration multiplied by time has units of metres per second, matching velocity. In the displacement relationship, each term has units of metres.

A sign check asks whether the direction matches the situation and the chosen positive direction. A reasonableness check asks whether the size makes sense. Near Earth's surface, an object's velocity changes by about

9.81\ \mathrm{m/s}

each second in the downward direction in this model. This estimate helps you judge whether a calculated speed or change in speed is plausible.

Do not confuse displacement with distance. Displacement includes direction and can be negative. Distance is the total path length and is not negative. The equations here use signed displacement, not total distance.

[a t]=\mathrm{m/s},\qquad [v_i t]=[a t^2]=\mathrm{m}

Check dimensions, signs, direction, and approximate size before reporting an answer.
Use displacement, not distance, in the signed constant-acceleration relationships.
State direction in words so the physical meaning is clear.

Worked example

1. Dropping a ball

A ball is released from rest and falls for

1.20\ \mathrm{s}

. Find its velocity at the end of that time. Use the free-fall model.

Set the system and direction
The system is the ball. Choose downward as positive. It is released from rest, so its initial velocity is zero. Gravity points in the positive direction.
$v_i=0\ \mathrm{m/s},\quad t=1.20\ \mathrm{s},\quad a=+9.81\ \mathrm{m/s^2}$
Select and substitute
The known time and acceleration connect to the final velocity through the constant-acceleration velocity relationship. The resulting unit is metres per second.
$v_f=v_i+at=(0\ \mathrm{m/s})+(9.81\ \mathrm{m/s^2})(1.20\ \mathrm{s})=11.772\ \mathrm{m/s}$
Round and check
The given time has three significant figures, so report three significant figures. The positive sign means downward. A speed increase of roughly $9.81\ \mathrm{m/s}$ in one second makes a result near $12\ \mathrm{m/s}$ after $1.20\ \mathrm{s}$ reasonable.
$v_f=11.8\ \mathrm{m/s}$

Answer: The ball's velocity is

11.8\ \mathrm{m/s}

downward.

Check: The answer has velocity units, points downward as expected, and is close to the estimated change of about

9.81\times1.20\approx11.8\ \mathrm{m/s}

.

Worked example

2. Tossing a ball upward

A ball is thrown vertically upward at

14.0\ \mathrm{m/s}

. How long does it take to reach its highest point? Use the free-fall model.

Set the system and direction
The system is the ball. Choose upward as positive. The initial velocity is positive, while gravity's acceleration is negative. At the highest point, the ball's vertical velocity is zero for an instant.
$v_i=+14.0\ \mathrm{m/s},\quad v_f=0\ \mathrm{m/s},\quad a=-9.81\ \mathrm{m/s^2}$
Rearrange for time
Use the velocity relationship because it includes the known initial velocity, final velocity, and acceleration. Solving for time gives the velocity change divided by acceleration.
$t=\frac{v_f-v_i}{a}=\frac{0\ \mathrm{m/s}-14.0\ \mathrm{m/s}}{-9.81\ \mathrm{m/s^2}}=1.427\ \mathrm{s}$
Round and check
The initial velocity has three significant figures, so report three. The time is positive. Gravity reduces the upward velocity by about $9.81\ \mathrm{m/s}$ each second, so reaching zero from $14.0\ \mathrm{m/s}$ takes about $1.43\ \mathrm{s}$ .
$t=1.43\ \mathrm{s}$

Answer: The ball takes

1.43\ \mathrm{s}

to reach its highest point.

Check: The units reduce to seconds. The positive time and zero velocity at the top fit the described motion; acceleration remains downward at the highest point.

Worked example

3. Finding the height from a fall

A rock is released from rest and reaches a speed of

19.6\ \mathrm{m/s}

downward. How far has it fallen? Use the free-fall model.

Set the system and direction
The system is the rock. Choose downward as positive. The rock starts from rest, and its final velocity is downward. Both its acceleration and its displacement during the fall are positive.
$v_i=0\ \mathrm{m/s},\quad v_f=+19.6\ \mathrm{m/s},\quad a=+9.81\ \mathrm{m/s^2}$
Use the relationship without time
Time is not given and is not needed. Use the relationship connecting velocity, acceleration, and displacement, then solve for displacement. Keep the squared velocity units in the substitution.
$\Delta d=\frac{v_f^2-v_i^2}{2a}=\frac{(19.6\ \mathrm{m/s})^2-(0\ \mathrm{m/s})^2}{2(9.81\ \mathrm{m/s^2})}=19.58\ \mathrm{m}$
Round and check
The supplied values have three significant figures. The positive displacement is downward. The result is reasonable: starting from rest, falling about $19.6\ \mathrm{m}$ under gravity produces a speed close to $19.6\ \mathrm{m/s}$ .
$\Delta d=19.6\ \mathrm{m}$

Answer: The rock has fallen

19.6\ \mathrm{m}

downward.

Check: The calculation has displacement units because velocity squared divided by acceleration gives metres. The direction and approximate speed-height relationship are consistent with free fall.

Common mistakes and how to avoid them

Using

+9.81\ \mathrm{m/s^2}

for gravity in every problem.

Correction: Assign the sign after choosing the positive direction. Gravity is negative when upward is positive and positive when downward is positive.

Assuming acceleration is zero at the highest point of an upward toss.

Correction: The velocity is momentarily zero there, but the free-fall acceleration is still downward.

Treating a negative velocity or displacement as an error.

Correction: A negative sign shows that the vector points opposite to the chosen positive direction. Interpret it using the coordinate choice.

Using distance in place of signed displacement.

Correction: The constant-acceleration relationships use displacement along the coordinate. Give its sign and direction.

Lesson summary

Free fall is modelled as motion in which gravity alone affects the object, with air resistance ignored.
Gravity has magnitude about $9.81\ \mathrm{m/s^2}$ near Earth's surface and points downward.
Choose a positive direction, assign signs consistently, and use a constant-acceleration relationship that includes the unknown.
Report SI units, sensible significant figures, and direction; then check units, sign, and reasonableness.

Check your understanding

Question 1

A ball is moving upward while in free fall. If upward is positive, which statement describes its acceleration?

It is positive because the ball is moving upward.
It is zero because the ball is not moving downward yet.
It is negative because gravity points downward.
It changes direction as the ball slows down.

Show answer and explanation

It is negative because gravity points downward.

Acceleration points in the direction of gravity throughout the free-fall motion. With upward chosen as positive, downward acceleration is negative.

Question 2

A stone is released from rest and falls for

2.00\ \mathrm{s}

. Taking downward as positive, what is its velocity after that time?

$+4.91\ \mathrm{m/s}$
$-19.6\ \mathrm{m/s}$
$+19.6\ \mathrm{m/s}$
$+9.81\ \mathrm{m/s}$

Show answer and explanation

+19.6\ \mathrm{m/s}

Using

v_f=v_i+at

gives

0+(9.81\ \mathrm{m/s^2})(2.00\ \mathrm{s})=+19.6\ \mathrm{m/s}

. The positive direction is downward.

Question 3

At the highest point of a ball's upward free-fall motion, which pair is correct?

Velocity is zero and acceleration is downward.
Velocity and acceleration are both zero.
Velocity is downward and acceleration is zero.
Velocity and acceleration are both upward.

Show answer and explanation

Velocity is zero and acceleration is downward.

The ball's vertical velocity is momentarily zero at the top, but gravity still causes downward acceleration.

Key terms

Acceleration: The change in velocity per unit time. It is a vector and has SI units of metres per second squared.
Displacement: The change in position along a chosen direction. It is a vector and can be positive or negative.
Free fall: Motion modelled as being affected only by gravity, with air resistance ignored.
Positive direction: The direction chosen to represent positive values for a coordinate.
Velocity: The rate of change of position in a stated direction. It is a vector measured in metres per second.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Physics (SPH3U), expectation C2.6. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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