Projectiles 1

1
Projectiles (1)
2
Vertical Motion
When an object is dropped, it accelerates
towards the ground.
The acceleration is cause by gravity.
Previously, we have met g
g
Where g 9.8 ms-2
- this is the acceleration of all bodies on earth
under gravity. (ignoring air resistance)
3
Constant Acceleration Equations
Conventions in Mechanics v - final velocity u -
initial velocity a - acceleration t time s -
displacement
In projectiles ... (body under free-fall) a
will always be -9.8 ms-2
It is negative, because in this topic we usually
measure things upwards. and objects fall
downwards.
4
The Constant Acceleration Equations
v u at
s ut 1/2 at2
v2 u2 2as
s 1/2 (vu) t
The equations used previously are used in
projectiles
5
A ball is dropped from a tower. It takes 4
seconds to hit the ground. How high is the tower?
4 seconds
Height ?
6
Using the Constant Acceleration Equations
s 1/2 (vu) t
v2 u2 2as
v u at
s ut 1/2 at2
A ball is dropped from a tower. It takes 4
seconds to hit the ground. How high it the tower?
What do I want to know? s - displacement
What do I know? a - acceleration u - initial
velocity t time
-9.8 ms-2
0 ms-1
4 sec.
Which equation involves these 4 ?
s ut 1/2 at2
s 0 x 4 1/2 x -9.8 x 42
s -4.9 x 16 -78.4 m
7
An object is thrown upwards at 20 ms-1. What is
its highest point? How long does it take to
reach this point?
20 ms-1
8
Using the Constant Acceleration Equations
s 1/2 (vu) t
v2 u2 2as
v u at
s ut 1/2 at2
An object is thrown upwards at 20 ms-1. What is
its highest point? How long does it take to
reach this point?
What do I know? v - final velocity u - initial
velocity a acceleration
What do I want to know? s - displacement t - time
0 ms-1
20 ms-1
-9.8 ms-2
v u at
Which equation involves 4 of these?
0 20 - 9.8t
t 20 / 9.8 2.04 sec.
9.8t 20
9
Using the Constant Acceleration Equations
s 1/2 (vu) t
v2 u2 2as
v u at
s ut 1/2 at2
An object is thrown upwards at 20 ms-1. What is
its highest point? How long does it take to
reach this point?
20.4 m above the ground
2.0 sec (1 d.p.)
What do I know? v - final velocity u - initial
velocity a acceleration t - time
What do I want to know? s - displacement
0 ms-1
20 ms-1
-9.8 ms-2
2.04 sec
s 1/2 (vu) t
Which equation involves 4 of these?
s 1/2 (020) x 2.04
s 10 x 2.04 20.4 m
10
An object is thrown upwards at 12 ms-1. Find the
total time in the air?
12 ms-1
11
Using the Constant Acceleration Equations
s 1/2 (vu) t
v2 u2 2as
v u at
s ut 1/2 at2
An object is thrown upwards at 12 ms-1. Find the
total time in the air?
What do I know? s - displacement u - initial
velocity a acceleration
What do I want to know? t - time
0 m
12 ms-1
-9.8 ms-2
s ut 1/2 at2
Which equation involves 4 of these?
0 12t 1/2 x -9.8 x t2
t(4.9t - 12) 0
0 12t - 4.9t2
t0 or 4.9t - 12 0 t 12 / 4.9 2.45 sec
4.9t2 - 12t 0
12
Projectile Motion
Objects thrown in a PROJECTILE MOTION fly in a
parabola
in Mechanics 1 all projectile motion - assumes
negligible mass - assumes no air resistance -
no wind - no spin
13
This is no good. Wind and air resistance will
effect the motion too much.
14
Freddie hitting a six
15
Constant Acceleration Equations in 2D
Given as vectors v - final velocity u - initial
velocity a - acceleration s - displacement As a
scaler t time
j
i
a will always be -9.8j ms-2
16
A football is kicked horizontally off a cliff at
10 ms-1
u - initial velocity ?
u 10i ms-1
What is the displacement through time?
s ut 1/2 at2
s 10i t 1/2 x -9.8j x t2
s 10t i - 4.9 t2 j
17
s 10t i - 4.9 t2 j
Time 0 1 2 3 4
5 Displacement
20i -19.6j
30i -44.1j
40i -78.4j
50i -122.5j
10i -4.9j
0
Parabolic Motion
18
Activity
s 1/2 (vu) t
v2 u2 2as
v u at
s ut 1/2 at2

• Page 122 of your Mechanics 1 book and answer
• exercise A