Volume 4: Mechanics 1
Vertical Motion under Gravity
If we drop a stone, it falls towards the Earth.
If we model the stone as a particle and ignore air
resistance, the only force acting on the stone is
its weight.
Weight is the force due to gravity.
The gravitational force produces an acceleration which
at the Earth’s surface is approximately
9·8 m s -2.
acceleration, a = g = 9·8 m s -2
To solve problems involving vertical motion under gravity,
we can use the equations of motion for constant
acceleration.
There are 2 types of problems
1) Object initially thrown up – hence it initially slows down
acc = -9.8ms-2. Make UP +ve
2) Object initially thrown down or dropped – hence it speeds
up
acc = +9.8ms-2 Make DOWN +ve
What happens initially is crucial – not whether it comes down
later.
e.g.1 A stone falls from rest from a point 2 m above
the ground. Modelling the stone as a particle and
ignoring air resistance, find its velocity as it hits the
ground.
Solution:
s, u, v, a, t
e.g.1 A stone falls from rest from a point 2 m above
the ground. Modelling the stone as a particle and
ignoring air resistance, find its velocity as it hits the
ground.
Solution:
s, u, v, a, t
e.g.1 A stone falls from rest from a point 2 m above
the ground. Modelling the stone as a particle and
ignoring air resistance, find its velocity as it hits the
ground.
Solution:
s, u, v, a, t
a = g = +9·8 m s -2 as initially dropped. Make DOWN +ve
e.g.1 A stone falls from rest from a point 2 m above
the ground. Modelling the stone as a particle and
ignoring air resistance, find its velocity as it hits the
ground.
Solution:
a = g = 9·8 m s -2
?
s, u, v, a, t
e.g.1 A stone falls from rest from a point 2 m above
the ground. Modelling the stone as a particle and
ignoring air resistance, find its velocity as it hits the
ground.
Solution:
a = g = 9·8 m s -2
?
s, u, v, a, t
e.g.1 A stone falls from rest from a point 2 m above
the ground. Modelling the stone as a particle and
ignoring air resistance, find its velocity as it hits the
ground.
Solution:
a = g = 9·8 m s -2
?
s, u, v, a, t
v = u + at
s = ut + 12 at 2
s = 12 (u + v)t
v2 = u2 + 2as
e.g.1 A stone falls from rest from a point 2 m above
the ground. Modelling the stone as a particle and
ignoring air resistance, find its velocity as it hits the
ground.
?
s, u, v, a, t
Solution:
a = g = 9·8 m s -2
We need the equation without t.
v2 = u2 + 2as



v2 = 0 + 2(9·8)(2)
v2 = 39·2
v = 6·26 ( 3 s.f. )
The velocity is 6·26 m s -1.
v = u + at
s = ut + 12 at 2
s = 12 (u + v)t
v2 = u2 + 2as
e.g.2 I throw a ball vertically upwards so that as it
leaves my hand it has a velocity of 14·7 m s -1.
Modelling the ball as a particle and ignoring air
resistance, find
(a) the time taken to reach the maximum height,
(b) the velocity and displacement after 2 s.
s, u, v, a, t
Solution:
(a)
As the ball will travel up and
down, we must be careful with
signs. Always draw a diagram.
u = 14·7
a = g = -9·8 m s -2 as initially thrown up. Make UP +ve
e.g.2 I throw a ball vertically upwards so that as it
leaves my hand it has a velocity of 14·7 m s -1.
Modelling the ball as a particle and ignoring air
resistance, find
(a) the time taken to reach the maximum height,
(b) the velocity and displacement after 2 s.
s, u, v, a, t
Solution:
(a)
At the maximum height, the ball
is neither going up nor down, so
the velocity is zero.
v=0
u = 14·7
e.g.2 I throw a ball vertically upwards so that as it
leaves my hand it has a velocity of 14·7 m s -1.
Modelling the ball as a particle and ignoring air
resistance, find
(a) the time taken to reach the maximum height,
(b) the velocity and displacement after 2 s.
Solution:
(a)
s, u, v, a, t ?
v=0
The ball is thrown up. What is the
acceleration ?
a = -9·8
Ans: The ball is slowing down, so
the acceleration is negative.
u = 14·7
Make UP +ve
s, u, v, a, t
v = u + at



?
 0 = 14·7 + (-9·8)t
9·8t = 14·7
t = 14·7
9·8
t = 1·5 s
v=0
a = -9·8
u = 14·7
Graph of
Ex
s=20t - at2
s=ut + at2
(b) Find the velocity and displacement after 2 s.
?
s, u, v, a, t
Since the ball has taken 1·5 s to
reach its greatest height, we
know that it will be
on its way down after 2 s.
We
not
needmotion
to change
( The
up do
and
down
is in one
the sign
of aclearer
.
straight line
but it’s
if we
draw the lines side by side. )
t=2
a = -9·8
u = 14·7
(b) Find the velocity and displacement after 2 s.
?
s, u, v, a, t
v = u + at


v = 14·7 + (-9·8)2
v = -4·9 m s -1
After 2 s the velocity is -4·9 m s -1.
( The ball is coming down. )
t=2
a = -9·8
u = 14·7
s = ut + 12 at 2


s = 14·7(2) + 12 (-9·8)(2) 2
s = 9·8 m
( After 2 s the ball is 9·8 m above the point where it
left my hand. )
e.g.2 A stone is thrown upwards with a velocity
of 5 m s -1 at the edge of a cliff and falls into the
sea 60 m vertically below the starting point.
Modelling the ball as a particle and ignoring air
resistance, find the length of time it is in the air.
In this question, assume the value of g is 10 m s -2
Solution:
s, u, v, a, t
u=5
a = -10
e.g.2 A stone is thrown upwards with a velocity
of 5 m s -1 at the edge of a cliff and falls into the
sea 60 m vertically below the starting point.
Modelling the ball as a particle and ignoring air
resistance, find the length of time it is in the air.
In this question, assume the value of g is 10 m s -2
Solution:
s, u, v, a, t ?
If we assume that the displacement,
s, is zero when the stone is thrown,
s will be negative below this level.
Make UP +ve
u=5
s=0
a = -10
s = -60
Solution:
s = ut + 12 at 2
s, u, v, a, t ?
-60 = 5t +
-12 = t - t 2
Divide by 5:

t 2 - t - 12 = 0
(t + 3)(t equation
- 4) = 0 so we must
This is a
quadratic
 t or
= -3
or the
+4 formula.
either factorise
use

u=5
s=0
1 (-10)t 2
2
a = -10
s = -60
WeThe
needstone
zerotakes
on one4 sside.
to reach the sea.
What meaning we can give to t = -3
Ans: A negative value represents time before
the stone was thrown.
e.g.4. A stone is dropped from rest into a well and
falls into the water 10 m below. Ignoring all
forces except the force due to gravity, find
the time taken until it hits the water.
Solution:
s = ut +
s, u, v, a, t ?
1 at 2
2
 10 = 0t + 12 9·8 t 2
In this question,
all=the
is
2
 10
4·9tmotion
downwards so, Make DOWN +ve
 10 = t 2
4·9
t = 1·43 s

( The negative root is not possible. )
u=0
a = 9·8
s = 10
EXERCISE
1. A ball is dropped from rest at the top of a tower
and hits the ground 2 s later. Modelling the ball
as a particle and ignoring air resistance, find the
height of the tower.
Solution:
s = ut +
?
s, u, v, a, t
1 at 2
2
s = 0t + 12 9·8(2) 2
s = 19·6
The tower is 19·6 m high.


u=0
a = 9·8
t=2
2. A ball is thrown upwards with a velocity of 12 m s -1
and travels in a straight line landing on the ground
1 m below its starting height.
Modelling the ball as a particle and ignoring air
resistance, find how fast it is moving as it hits
the ground.
?
Solution:
s, u, v, a, t
v2 = u2 + 2as
 v 2 = 12 2 + 2 (-9·8)(-1)
 v 2 = 163·6
 v = 12·8 m s -1
a = -9·8
u = 12
s = -1
Past M1 question
Answers
Notes page 27
SUMMARY
 For vertical motion where the only force is the
weight ( the force due to gravity ) we can use the
equations of motion for constant acceleration.
 The acceleration due to gravity has a magnitude of
9·8 m s -2 close to the surface of the Earth.
 We draw a sketch of the motion to avoid sign
errors.
 For motion that is all downwards, we usually take
the positive axis downwards so g = +9·8. Otherwise,
the positive axis is up and g = -9·8
 There is no need to separate the parts of the
motion where the body is moving up and down.
The summary page follows in a form suitable for
photocopying.
TEACH A LEVEL MATHS – MECHANICS 1
VERTICAL MOTION UNDER GRAVITY
Summary

For vertical motion where the only force is the weight ( the force due to
gravity ) we can use the equations of motion for constant acceleration.

The acceleration due to gravity has a magnitude of
surface of the Earth.

We draw a sketch of the motion to avoid sign errors.

For motion that is all downwards, we usually take the positive axis
downwards so g = +9·8. Otherwise, the positive axis is up and g =

There is no need to separate the parts of the motion where the body is
moving up and down.
9·8
m s -2 close to the
-9·8.