3.3.3 Weight and motion under gravity

What you'll learn

How mass, weight and gravitational acceleration are connected.
Why the value of gravitational acceleration depends on location.
How to model vertical motion using constant-acceleration equations.
How to handle objects that are dropped, projected upwards or moving downwards.

Mass, weight and gravity

Mass

Mass measures the amount of matter in an object. Its SI unit is the kilogram, kg.

An object's mass does not change simply because it moves to a different location. For example, an astronaut has the same mass on Earth and on the Moon.

Weight

Weight is the gravitational force acting on an object. As it is a force, its SI unit is the newton, N.

Definition

Weight

For an object of mass $m$ , its weight $W$ is

W = m g,

where $g$ is the magnitude of the gravitational acceleration at that location.

Weight acts vertically downwards, towards the centre of the Earth or other astronomical body.

Common Mistake

Mass is not weight

Mass is measured in kilograms, while weight is measured in newtons. A statement such as “the weight is 5 kg” confuses two different quantities.

Example

Calculating weight

A suitcase has mass 18 kg. Calculate its weight where $s−2g=9.8\text{ m s}^{-2}$ .

Use the relationship $W = m g$ .
Substitute the mass and gravitational acceleration:
$W=18×9.8.W=18\times 9.8.$
Therefore, the suitcase has weight $N176.4\text{ N}$ , acting vertically downwards.

Gravitational acceleration

An object falling under gravity accelerates because its weight produces a resultant downward force.

Definition

Gravitational acceleration

The symbol $g$ represents the magnitude of the acceleration due to gravity. Near the Earth's surface, its usual A-Level value is

s−2.g=9.8\text{ m s}^{-2}.

This means that, when air resistance is ignored, the object's downward velocity increases by approximately 9.8 m s⁻¹ every second.

Questions may specify different levels of accuracy, such as 9.81 m s⁻² or 10 m s⁻². Use the value given in the question. If none is given, use 9.8 m s⁻².

Although $g$ is often treated as constant in mechanics models, it is not a universal constant. Its value depends on location, including altitude and the astronomical body involved. You are not required to use the inverse square law for gravitation in this topic.

Key Idea

The standard model

Near the Earth's surface, an object moving under gravity alone is modelled as having constant acceleration of magnitude $g$ , directed vertically downwards.

Free fall and modelling assumptions

An object is in free fall when gravity is the only force acting on it. In the standard model, air resistance is ignored.

This has an important consequence: all freely falling objects at the same location have the same acceleration, regardless of their mass.

Using Newton's second law confirms this. For an object of mass $m$ falling freely, its weight is the resultant force:

F=ma,mg=ma,a=g.\begin{aligned} F&=ma,\\ mg&=ma,\\ a&=g. \end{aligned}

The mass cancels, so a heavier object does not have a greater free-fall acceleration.

Force and motion diagrams for free fall and upward-positive vertical motion

Common Mistake

When the constant-acceleration model is unsuitable

If air resistance is significant, gravity is not the only force acting. The object's acceleration is then not necessarily constant or equal to $g$ , so the usual constant-acceleration equations may not apply over the whole motion.

Choosing a positive direction

Velocity, displacement and acceleration are vector quantities, so their signs depend on the direction chosen as positive.

You may choose either upwards or downwards as positive. State your choice clearly and use it consistently.

Upwards as positive

If upwards is positive, then gravitational acceleration is negative:

s−2.a=-g=-9.8\text{ m s}^{-2}.

An upward velocity is positive, while a downward velocity is negative.

Downwards as positive

If downwards is positive, then

s−2.a=g=9.8\text{ m s}^{-2}.

A downward displacement and velocity are then positive.

Tip

Choose the convenient direction

For an object projected upwards, choosing upwards as positive often makes the initial velocity positive. For an object dropped down a shaft, choosing downwards as positive usually keeps the main quantities positive.

Common Mistake

Using positive g automatically

The value 9.8 is the magnitude of gravitational acceleration. The signed acceleration is $s−2+9.8\text{ m s}^{-2}$ or $s−2-9.8\text{ m s}^{-2}$ according to your chosen positive direction.

Constant-acceleration equations

When an object moves in a straight vertical line under gravity alone, its acceleration is constant. You can therefore use the SUVAT equations:

v=u+at,s=ut+12at2,v2=u2+2as,s=12(u+v)t,s=vt−12at2.\begin{aligned} v&=u+at,\\ s&=ut+\frac{1}{2}at^2,\\ v^2&=u^2+2as,\\ s&=\frac{1}{2}(u+v)t,\\ s&=vt-\frac{1}{2}at^2. \end{aligned}

Here:

$s$ is displacement from the chosen starting point, in m;
$u$ is initial velocity, in m s⁻¹;
$v$ is final velocity, in m s⁻¹;
$a$ is constant acceleration, in m s⁻²;
$t$ is elapsed time, in s.

Displacement is the signed change in position. It is not necessarily the same as total distance travelled.

An object dropped from rest

“Dropped” or “released from rest” means that the initial velocity is zero, so $u = 0$ .

Example

Dropping an object from a tower

A stone is dropped from rest from a tower and falls 30 m. Ignore air resistance. Find its speed immediately before it reaches the ground.

Choose downwards as positive. Then $u = 0$ , $ms=30\text{ m}$ and $s−2a=9.8\text{ m s}^{-2}$ . Time is not required, so use $v^2=u^2+2as$ .
Substitute the known quantities:
$v2=02+2×9.8×30=588.v^2=0^2+2\times 9.8\times 30=588.$
Since the question asks for speed, take the positive square root:
$s−1.v=\sqrt{588}=14\sqrt{3}\text{ m s}^{-1}.$

When an equation gives $v=±⋯v=\pm\sqrt{\cdots}$ , select the sign that matches the actual direction of motion. If the question asks for speed, give the non-negative magnitude.

An object projected vertically upwards

When an object is projected upwards, gravity slows it down as it rises. At its greatest height, its instantaneous velocity is zero.

Key Idea

At maximum height

At the highest point, $v = 0$ , but the acceleration is still directed downwards and has magnitude $g$ . The object has not lost its acceleration.

Example

Finding maximum height and time

A ball is projected vertically upwards at 14 m s⁻¹. Find the time taken to reach its greatest height and the height gained.

Choose upwards as positive, giving $s−1u=14\text{ m s}^{-1}$ , $s−2a=-9.8\text{ m s}^{-2}$ and $v = 0$ at maximum height. From $v = u + a t$ :
$0 = 14 - 9.8 t,$
so $st=\frac{10}{7}\text{ s}$ .
To find the height gained without introducing another unknown, use $v^2=u^2+2as$ :
$0^2=14^2+2(-9.8)s.$
Solving gives
$m.s=\frac{196}{19.6}=10\text{ m}.$

Common Mistake

Zero velocity does not mean zero acceleration

At the highest point, the ball is momentarily stationary, but gravity continues to act. Its acceleration remains $s−2-9.8\text{ m s}^{-2}$ when upwards is positive.

Returning to the starting height

If an object returns to the same height from which it was projected, its overall displacement is zero.

In the model with constant gravity and no air resistance:

the time rising equals the time falling back to the starting height;
the speed on return equals the initial speed;
the return velocity has the opposite sign to the initial velocity.

Example

Returning to the point of projection

A particle is projected vertically upwards at 19.6 m s⁻¹. Find when it returns to its starting point and its velocity then.

Choose upwards as positive. At the return point, $s = 0$ , while $s−1u=19.6\text{ m s}^{-1}$ and $s−2a=-9.8\text{ m s}^{-2}$ . Use $s=ut+12at2s=ut+\frac12at^2$ :
$0=19.6t-4.9t^2.$
Factorise:
$0 = 4.9 t (4 - t),$
giving $t = 0$ or $st=4\text{ s}$ . The value $t = 0$ describes the launch, so the particle returns after 4 s.
Find the return velocity using $v = u + a t$ :
$s−1.v=19.6-9.8(4)=-19.6\text{ m s}^{-1}.$
The negative sign means the particle is moving downwards.

Interpreting multiple solutions

A quadratic equation for time can produce two valid positive answers. This usually means that the object passes the same height once while rising and once while falling.

Do not reject a second time automatically. Interpret each solution using the physical motion.

Tip

Check the meaning of your answer

A negative velocity may be completely correct: it indicates motion opposite to your chosen positive direction. By contrast, a negative value of time is usually outside the modelled interval and should normally be rejected.

Exam technique

In the exam

Draw a simple vertical diagram, choose a positive direction and assign signs to $u$ , $v$ , $a$ and $s$ consistently.
Translate phrases carefully: “dropped” means $u = 0$ , “greatest height” means $v = 0$ , and “returns to its starting point” means $s = 0$ .
Select the SUVAT equation containing the quantities you know and the one you need, substitute with units, and interpret signs or multiple solutions in context.
Use the stated value of $g$ and avoid rounding until the final answer.

Self review

Check yourself

What is the difference between an object's mass and its weight?
If upwards is positive, what signs should displacement, velocity and acceleration have while an object is falling?
A ball is projected upwards and later passes the launch point again. What are its displacement and direction of velocity at that instant?

What you'll learn

How mass, weight and gravitational acceleration are connected.
Why the value of gravitational acceleration depends on location.
How to model vertical motion using constant-acceleration equations.
How to handle objects that are dropped, projected upwards or moving downwards.

Mass, weight and gravity

Mass

Mass measures the amount of matter in an object. Its SI unit is the kilogram, kg.

An object's mass does not change simply because it moves to a different location. For example, an astronaut has the same mass on Earth and on the Moon.

Weight

Weight is the gravitational force acting on an object. As it is a force, its SI unit is the newton, N.

Definition

Weight

For an object of mass $m$ , its weight $W$ is

W = m g,

where $g$ is the magnitude of the gravitational acceleration at that location.

Weight acts vertically downwards, towards the centre of the Earth or other astronomical body.

Common Mistake

Mass is not weight

Mass is measured in kilograms, while weight is measured in newtons. A statement such as “the weight is 5 kg” confuses two different quantities.

Example

Calculating weight

A suitcase has mass 18 kg. Calculate its weight where $s−2g=9.8\text{ m s}^{-2}$ .

Use the relationship $W = m g$ .
Substitute the mass and gravitational acceleration:
$W=18×9.8.W=18\times 9.8.$
Therefore, the suitcase has weight $N176.4\text{ N}$ , acting vertically downwards.

Gravitational acceleration

An object falling under gravity accelerates because its weight produces a resultant downward force.

Definition

Gravitational acceleration

The symbol $g$ represents the magnitude of the acceleration due to gravity. Near the Earth's surface, its usual A-Level value is

s−2.g=9.8\text{ m s}^{-2}.

This means that, when air resistance is ignored, the object's downward velocity increases by approximately 9.8 m s⁻¹ every second.

Questions may specify different levels of accuracy, such as 9.81 m s⁻² or 10 m s⁻². Use the value given in the question. If none is given, use 9.8 m s⁻².

Key Idea

The standard model

Near the Earth's surface, an object moving under gravity alone is modelled as having constant acceleration of magnitude $g$ , directed vertically downwards.

Free fall and modelling assumptions

An object is in free fall when gravity is the only force acting on it. In the standard model, air resistance is ignored.

This has an important consequence: all freely falling objects at the same location have the same acceleration, regardless of their mass.

Using Newton's second law confirms this. For an object of mass $m$ falling freely, its weight is the resultant force:

F=ma,mg=ma,a=g.\begin{aligned} F&=ma,\\ mg&=ma,\\ a&=g. \end{aligned}

The mass cancels, so a heavier object does not have a greater free-fall acceleration.

Force and motion diagrams for free fall and upward-positive vertical motion

Common Mistake

When the constant-acceleration model is unsuitable

Choosing a positive direction

Velocity, displacement and acceleration are vector quantities, so their signs depend on the direction chosen as positive.

You may choose either upwards or downwards as positive. State your choice clearly and use it consistently.

Upwards as positive

If upwards is positive, then gravitational acceleration is negative:

s−2.a=-g=-9.8\text{ m s}^{-2}.

An upward velocity is positive, while a downward velocity is negative.

Downwards as positive

If downwards is positive, then

s−2.a=g=9.8\text{ m s}^{-2}.

A downward displacement and velocity are then positive.

Tip

Choose the convenient direction

Common Mistake

Using positive g automatically

The value 9.8 is the magnitude of gravitational acceleration. The signed acceleration is $s−2+9.8\text{ m s}^{-2}$ or $s−2-9.8\text{ m s}^{-2}$ according to your chosen positive direction.

Constant-acceleration equations

When an object moves in a straight vertical line under gravity alone, its acceleration is constant. You can therefore use the SUVAT equations:

v=u+at,s=ut+12at2,v2=u2+2as,s=12(u+v)t,s=vt−12at2.\begin{aligned} v&=u+at,\\ s&=ut+\frac{1}{2}at^2,\\ v^2&=u^2+2as,\\ s&=\frac{1}{2}(u+v)t,\\ s&=vt-\frac{1}{2}at^2. \end{aligned}

Here:

$s$ is displacement from the chosen starting point, in m;
$u$ is initial velocity, in m s⁻¹;
$v$ is final velocity, in m s⁻¹;
$a$ is constant acceleration, in m s⁻²;
$t$ is elapsed time, in s.

Displacement is the signed change in position. It is not necessarily the same as total distance travelled.

An object dropped from rest

“Dropped” or “released from rest” means that the initial velocity is zero, so $u = 0$ .

Example

Dropping an object from a tower

A stone is dropped from rest from a tower and falls 30 m. Ignore air resistance. Find its speed immediately before it reaches the ground.

Choose downwards as positive. Then $u = 0$ , $ms=30\text{ m}$ and $s−2a=9.8\text{ m s}^{-2}$ . Time is not required, so use $v^2=u^2+2as$ .
Substitute the known quantities:
$v2=02+2×9.8×30=588.v^2=0^2+2\times 9.8\times 30=588.$
Since the question asks for speed, take the positive square root:
$s−1.v=\sqrt{588}=14\sqrt{3}\text{ m s}^{-1}.$

When an equation gives $v=±⋯v=\pm\sqrt{\cdots}$ , select the sign that matches the actual direction of motion. If the question asks for speed, give the non-negative magnitude.

An object projected vertically upwards

When an object is projected upwards, gravity slows it down as it rises. At its greatest height, its instantaneous velocity is zero.

Key Idea

At maximum height

At the highest point, $v = 0$ , but the acceleration is still directed downwards and has magnitude $g$ . The object has not lost its acceleration.

Example

Finding maximum height and time

A ball is projected vertically upwards at 14 m s⁻¹. Find the time taken to reach its greatest height and the height gained.

Choose upwards as positive, giving $s−1u=14\text{ m s}^{-1}$ , $s−2a=-9.8\text{ m s}^{-2}$ and $v = 0$ at maximum height. From $v = u + a t$ :
$0 = 14 - 9.8 t,$
so $st=\frac{10}{7}\text{ s}$ .
To find the height gained without introducing another unknown, use $v^2=u^2+2as$ :
$0^2=14^2+2(-9.8)s.$
Solving gives
$m.s=\frac{196}{19.6}=10\text{ m}.$

Common Mistake

Zero velocity does not mean zero acceleration

At the highest point, the ball is momentarily stationary, but gravity continues to act. Its acceleration remains $s−2-9.8\text{ m s}^{-2}$ when upwards is positive.

Returning to the starting height

If an object returns to the same height from which it was projected, its overall displacement is zero.

In the model with constant gravity and no air resistance:

the time rising equals the time falling back to the starting height;
the speed on return equals the initial speed;
the return velocity has the opposite sign to the initial velocity.

Example

Returning to the point of projection

A particle is projected vertically upwards at 19.6 m s⁻¹. Find when it returns to its starting point and its velocity then.

Choose upwards as positive. At the return point, $s = 0$ , while $s−1u=19.6\text{ m s}^{-1}$ and $s−2a=-9.8\text{ m s}^{-2}$ . Use $s=ut+12at2s=ut+\frac12at^2$ :
$0=19.6t-4.9t^2.$
Factorise:
$0 = 4.9 t (4 - t),$
giving $t = 0$ or $st=4\text{ s}$ . The value $t = 0$ describes the launch, so the particle returns after 4 s.
Find the return velocity using $v = u + a t$ :
$s−1.v=19.6-9.8(4)=-19.6\text{ m s}^{-1}.$
The negative sign means the particle is moving downwards.

Interpreting multiple solutions

A quadratic equation for time can produce two valid positive answers. This usually means that the object passes the same height once while rising and once while falling.

Do not reject a second time automatically. Interpret each solution using the physical motion.

Tip

Check the meaning of your answer

Exam technique

In the exam

Draw a simple vertical diagram, choose a positive direction and assign signs to $u$ , $v$ , $a$ and $s$ consistently.
Translate phrases carefully: “dropped” means $u = 0$ , “greatest height” means $v = 0$ , and “returns to its starting point” means $s = 0$ .
Select the SUVAT equation containing the quantities you know and the one you need, substitute with units, and interpret signs or multiple solutions in context.
Use the stated value of $g$ and avoid rounding until the final answer.

Self review

Check yourself

What is the difference between an object's mass and its weight?
If upwards is positive, what signs should displacement, velocity and acceleration have while an object is falling?
A ball is projected upwards and later passes the launch point again. What are its displacement and direction of velocity at that instant?

What you'll learn

Mass, weight and gravity

Mass

Weight

Gravitational acceleration

Free fall and modelling assumptions

Choosing a positive direction

Upwards as positive

Downwards as positive

Constant-acceleration equations

An object dropped from rest

An object projected vertically upwards

Returning to the starting height

Interpreting multiple solutions

Review 3.3.3 Weight and motion under gravity by teaching Genie

3.3.3 Weight and motion under gravity

What you'll learn

Mass, weight and gravity

Mass

Weight

Gravitational acceleration

Free fall and modelling assumptions

Choosing a positive direction

Upwards as positive

Downwards as positive

Constant-acceleration equations

An object dropped from rest

An object projected vertically upwards

Returning to the starting height

Interpreting multiple solutions

Review 3.3.3 Weight and motion under gravity by teaching Genie

3.3.3 Weight and motion under gravity

What you'll learn

Mass, weight and gravity

Mass

Weight

Gravitational acceleration

Free fall and modelling assumptions

Choosing a positive direction

Upwards as positive

Downwards as positive

Constant-acceleration equations

An object dropped from rest

An object projected vertically upwards

Returning to the starting height

Interpreting multiple solutions

Review 3.3.3 Weight and motion under gravity by teaching Genie

3.3.3 Weight and motion under gravity Revision Guide

3.3.3 Weight and motion under gravity

What you'll learn

Mass, weight and gravity

Mass

Weight

Gravitational acceleration

Free fall and modelling assumptions

Choosing a positive direction

Upwards as positive

Downwards as positive

Constant-acceleration equations

An object dropped from rest

An object projected vertically upwards

Returning to the starting height

Interpreting multiple solutions

Review 3.3.3 Weight and motion under gravity by teaching Genie

3.3.3 Weight and motion under gravity Revision Guide