Transcript 9.1

Fields and Forces
Topic 9.1 Projectile Motion
Components of Motion
When a body is in free motion, (moving
through the air without any forces apart from
gravity and air resistance), it is called a
projectile
Normally air resistance is ignored so the only
force acting on the object is the force due to
gravity
This is a uniform force acting downwards
Therefore if the motion of the projectile is
resolved into the vertical and horizontal
components
The horizontal component will be unaffected
as there are no forces acting on it
The vertical component will be accelerated
downwards by the force due to gravity
These two components can be
considered as independent factors in
the motion of a projectile in a uniform
field
In the absence of air resistance the
path taken by any projectile is parabolic
Solving Problems
In solving problems it is necessary to
consider the 2 components
independently
Therefore the horizontal motion it is
necessary to use the equation
speed = distance
time
Where speed is the horizontal
component of the velocity
Therefore the vertical motion it is
necessary to use the kinematic
equations for uniform acceleration
i.e. Using the s.u.v.a.t equations
Where u and v are the initial and final
vertical components of the velocity
Example
A ball is kicked at an angle of 40.0o with a
velocity of 10.0 ms-1. Taking g = 10 ms–2.
How far does it travel horizontally?
10ms-1
40o
To be able to calculate the horizontal
distance we need to know the
horizontal speed, and the time.
The horizontal distance is easy to
calculate by resolving the velocity
10.0 sin 40.0o
10.0ms-1
40.0o
10.0 cos 40.0o
However, to calculate the time we will
need to use the vertical component and
the s.u.v.a.t. Equations
s=?
u = 10.0 sin 40.0o ms-1
v=?
a = -10.0 ms-2 (Up is positive, therefore
acceleration here is negative)
t=?
We only have 2 of the values when we
need three to find any other
However, if we ignore air resistance,
then the final vertical component of the
velocity will be equal and opposite of
the initial component
i.e. v = -10.0 sin 40.0o ms-1
Looking at the equations for uniform
acceleration, we need an equation that
links u, v, a and t.
v = u + at
Rearranging to make t the subject
t=v–u
a
Substitute in
t = -10.0 sin 40.0o – 10.0 sin 40.0o
-10
t = 1.286 seconds
Now returning to the horizontal
components
Using speed = distance
time
Rearranging distance = speed x time
Distance = 10.0 cos 40.0o x 1.286
Distance = 9.851 = 9.9 metres
Using the Conservation of
Energy
In some situations the use of the
conservation of energy can be a much
simpler method than using the kinematic
equations
Solving projectile motion problems makes use
of the fact that Ek + Ep = constant at every
point in the objects flight (assuming no loss
of energy due to friction)
Example
A ball is projected at 25.0 ms-1 at an
angle of 40.00 to the horizontal. The ball
is released 2.00m above the ground.
Taking g = 10.0 ms-2. Find the
maximum height it reaches.
Solution
B
H
25.0 ms-1
A
2.0m
v = vhorizontal
Total energy at A is given by
Ek + Ep = ½ m (25.0)2 + mg x 2.0
=312.5m + 20m
= 332.5m
Next, to find the total energy at B we
need to know the velocity at B, which is
given by the horizontal component of
the velocity at A
Total energy at B is given by
Ek+ Ep = ½ m (25.0 cos 40o)2 + mg x H
=183.38m+ 10mH
Then using the conservation of energy
Equating the 2 equations
332.5m = 183.38m + 10mH
332.5 = 183.38 + 10H
332.5 – 183.38 = 10H
10H = 149.12
H = 14.912 = 14.9m