Transcript Projectile Motion

Projectile Motion
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Projectile Motion
9.1
Projectile motion (AHL)
9.1.1
State the independence of the vertical and the horizontal
components of velocity for a projectile in a uniform field.
9.1.2
Describe and sketch the trajectory of projectile motion as parabolic
in the absence of air resistance.
9.1.3
Describe qualitatively the effect of air resistance on the trajectory of
a projectile.
9.1.4
Solve problems on projectile motion.
Amazing facts!
If a gun is fired horizontally, and at the
same time a bullet is dropped from the
same height. They both hit the ground at
the same time.
Amazing facts!
Amazing facts!
Amazing facts!
Amazing facts!
Amazing facts!
Why?
Vertical and horizontal
Their vertical motion can be considered
separate from their horizontal motion.
Vertical and horizontal
independent of each other
Vertically, they both have zero initial
velocity and accelerate downwards at 9.8
m.s-2. The time to fall the same vertical
distance is therefore the same.
Horizontally velocity remains constant
Projectiles (Half Trajectory)
An object projected sideways through the air will follow a curved
trajectory.
horizontal motion
(steady speed)
VH 
vertical motion
DH
t
accelerates downwards at -9.8 ms-2
The horizontal and vertical motions should be treated separately.
Time is the only quantity common to both.
At any point in its trajectory, the velocity of a projectile has two
components.
• one vertical, VV
• the other horizontal, VH
The resultant velocity is found drawing a vector diagram and add the
vectors together, TIP to TAIL.
Vector Diagram
horizontal velocity
resultant/actual
velocity
vertical
velocity
vh
This is an example of a
‘half-trajectory.’
vv
GREEN – actual motion
RED – vertical motion
BLUE – horizontal motion
Watch that dog!
Imagine a dog being kicked horizontally off the
top of a cliff (with an initial velocity vh).
vh
Parabola
Assuming that there is negligible air
resistance, he falls in the path of a parabola.
Parabola
Parabola
Why?
Why a parabola?
We can consider his motion to be the sum of
his horizontal motion and vertical motion.
vh
We can treat these separately
Horizontal motion
Assuming no air resistance, there are no
horizontal forces.
vh
This means horizontally
the dog moves with
constant speed vh
Horizontal distance travelled (x) = vht
Vertical motion
Assuming no air resistance, there is constant
force downwards (=mg).
This means vertically the
dog moves with constant
acceleration g = 9.8 m.s-2
Vertical distance travelled (y) = uvt + ½gt2
Parabolic motion
Since y = ½gt2 (if u = 0) and x = vht,
y = ½gx2/vh2 which you may (!) recognise as
the formula of a parabola.
Another piece of ultra
cool physics!
Example
A ball is kicked horizontally off an embankment, with a velocity of 30
ms-1.
It lands 24 m from the base of the embankment.
(a)
Calculate how long the ball was in flight.
30 ms-1
24 m
VH 
DH
t
30 
24
t
t
24
30
t  0.8 s
common to
horizontal and
vertical motions
(b)
Calculate the horizontal velocity just before hitting the ground.
Horizontal
Vertical
u  30 ms -1
a  -9.8 ms -2
t  0.8 s
u  0 ms -1
s  24 m
t  0.8 s
a  0 ms-2
travels
horizontally at
steady speed –
no acceleration
horizontally
Horizontal
v  u  at
 30  0  0.8
v  30 ms -1
acted upon by
gravity
not initially
falling down, so
speed of zero in
vertical direction
(c)
Calculate the vertical velocity just before hitting the ground.
Vertical
v  u  at
 0  - 9.8 0.8
v  -7.84 ms -1
(d)
means 7.84 ms-1 downwards
How high is the embankment?
Vertical
s  ut 
a  -9.8 ms -2
t  0.8 s
u  0 ms -1
v  -7.84 ms -1
1 2
at
2
 0  0.8 
s  3.14 m
1
- 9.8  0.82
2


means ball fell through
distance of 3.14 m
so height of the embankment is 3.14 m
(e)
Calculate the resultant velocity of the ball, just before hitting
the ground.
30 ms-1
θ
velocity
-7.8 ms-1
Size
By Pythagoras:
a2  b2  c2
resultant velocity2  302  - 7.842
 900  61.5
resultant velocity 961.5
 31 ms -1
Direction
cos θ 
adj
hyp
30
31
θ  cos1 0.97
cos θ 
30 ms-1
θ
velocity
-7.8 ms-1
θ  14.6
resultant velocity 31 ms-1 at angle of 14.6 below horizon
Q1.
A ball is kicked off a cliff with a horizontal speed of 16 ms-1.
The ball hits the ground 2.2 s later.
(a)
Calculate the height of the cliff.
(b)
Calculate the distance between the foot of the
35.2 m
cliff and where the ball lands.
(c)
Calculate the vertical component of the balls
velocity just before it hits the ground. 21.6 ms-1
(d)
Calculate the balls velocity as it hits the
ground.
26.9 ms-1 at angle
23.7 m
of 53.5° below
horizon
You may want to draw a diagram to help you get started !!!
Q2.
A ball is kicked off a cliff with a horizontal speed of 22 ms-1.
the ball hits the ground 1.5 s later.
11 m
(a)
Calculate the height of the cliff.
(b)
Calculate the horizontal distance from the foot
33 m
of the cliff, to where the ball lands.
(c)
Calculate the vertical component of the balls
14.7 ms-1
velocity as it hits the ground.
(d)
Calculate the balls actual velocity as it hits the
ground.
26.5 ms-1 at angle of
34° below horizon
You may want to draw a diagram to help you get started !!!
Example
5 m.s-1
30 m
A dog is kicked off the
top of a cliff with an initial
horizontal velocity of 5
m.s-1. If the cliff is 30 m
high, how far from the
cliff bottom will the dog
hit the ground?
Example
Looking at vertical motion
first:
5 m.s-1
30 m
u = 0, a = 9.8 m.s-2, s = 30 m, t = ?
s = ut + ½at2
30 = ½ x 9.8 x t2
t2 = 6.1
t = 2.47 s
The dog hits the ground after
2.47 seconds (yes!)
Example
Now look at horizontal motion:
5 m.s-1
Constant speed (horizontally) = 5 m.s-1
Time of fall = 2.47 seconds
Horizontal distance travelled = speed x time
Horizontal distance travelled = 5 x 2.47
30 m
= 12.4 m
The dog hits the ground 12.4
metres from the base of the
cliff
Parabola
12.4 metres
What is the dog’s speed as he hits
the ground?
5 m.s-1
30 m
To answer this it is
easier to think in terms of
the dog’s total energy
(kinetic and potential)
What is the dog’s speed as he hits
the ground?
Total energy at top = ½mv2 + mgh
5 m.s-1
Total energy = ½m(5)2 + mx9.8x30
Total energy = 12.5m + 294m = 306.5m
30 m
What is the dog’s speed as he hits
the ground?
At the bottom, all the potential energy has
been converted to kinetic energy. All the dog’s
energy is now kinetic.
energy = ½mv2
V=?
What is the dog’s speed as he hits
the ground?
energy at top = energy at bottom
306.5m = ½mv2
306.5 = ½v2
613 = v2
V = 24.8 m.s-1
(Note that this is the dog’s
speed as it hits the ground,
not its velocity.
v = 24.8 m.s-1
Projectiles (Full Trajectory)
A projectile does not need to be an object falling, but could be an
object fired at angle to the horizontal.
θ
The subsequent motion would be
max height
If air resistance is ignored, the trajectory has an axis of symmetry
about the mid point (maximum height).
So the time taken to reach the maximum height is the same as the
time taken to fall back to the ground.
Various calculations can be made, but firstly, the initial velocity must
be split into its horizontal and vertical components.
Horizontal
Vertical
a = 0 ms-2
a = -9.8 ms-2
Starting with non-horizontal motion
Woof!
(help)
Starting with non-horizontal motion
25 m.s-1
30°
Starting with non-horizontal motion
1. Split the initial velocity into vertical and
horizontal components
vh = 25cos30°
vv = 25sin30°
25 m.s-1
30°
Starting with non-horizontal motion
2. Looking at the vertical motion, when the
dog hits the floor, displacement = 0
Initial vertical velocity = vv = 25sin30°
Acceleration = - 9.8 m.s-2
25 m.s-1
30°
Starting with non-horizontal motion
3. Using s = ut + ½at2
0 = 25sin30°t + ½(-9.8)t2
0 = 12.5t - 4.75t2
0 = 12.5 – 4.75t
4.75t = 12.5
t = 12.5/4.75 = 2.63 s
25 m.s
-1
30°
Starting with non-horizontal motion
4. Looking at horizontal motion
Ball in flight for t = 2.63 s travelling with
constant horizontal speed of
vh = 25cos30° = 21.7 m.s-1.
Distance travelled = vht = 21.7x2.63 = 57.1m
30°
57.1m
Starting with non-horizontal motion
5. Finding maximum height? Vertically;
v = 0, u = 25sin30°, t = 2.63/2
s = (u + v)t = 12.5x1.315 = 8.2m
2
2
30°
Starting with non-horizontal motion
6. Don’t forget some problems can also be
answered using energy.
30°
Starting with non-horizontal motion
6. Don’t forget some problems can also be
answered using energy.
As dog is fired total energy = ½m(25)2
25 m.s-1
30°
Starting with non-horizontal motion
6. At the highest point,
total energy = KE + GPE
=½m(25cos30°)2 + mgh
As dog is fired total energy = ½m(25)2
30°
Starting with non-horizontal motion
6. So ½m(25cos30°)2 + mgh = ½m(25)2
½(21.65)2 + 9.8h = ½(25)2
234.4 + 9.8h = 312.5
9.8h = 78.1
h = 8.0 m
30°