### Mathematically, how the projectile works

The AP Annual National
Conference
July 19, 2013
Ahmed Salama
P.C.C.C. College
[email protected]
1
AMTNJ Annual
Conference
October 24-25, 2014
Use trigonometric functions and
calculus skills to determine the
range and final velocity of a
projectile based in knowing its
initial velocity and maximum height.
This activity provides a good
example of connecting physics and
mathematics.
3
Mathematically, how the
projectile works !!
4
Rocket 2:
60
Rocket 1:
Range VS Theoretical Range at 40˚
60
40
40
30
Range
Distance
50
Distance
50
Range VS Theoretical Range at 40˚
30
Range
Theoretical Range
Theoretical Range
20
20
10
10
0
0
40
60
80
Pressure
100
40
60
80
Pressure
Compare rocket 1 to rocket 2 in range
and theoretical range at 400 angle
Rocket 1= .073 kg and Rocket 2 = .086 kg
100
Rocket 2:
Rocket 1:
Range VS Theoretical Range at 45˚
60
Range VS Theoretical Range at 45˚
60
50
50
40
30
Range
Theoretical Range
Distance
Distance
40
30
Theoretical Range
20
20
10
10
0
0
40
60
80
Pressure
100
Range
40
60
80
Pressure
Compare rocket 1 to rocket 2 in range
and theoretical range at 450 angle
Rocket 1= .073 kg and Rocket 2 = .086 kg
100
Rocket 1:
Rocket 2:
Range VS Theoretical Range at 50˚
Range VS Theoretical Range at 50˚
60
45
40
50
35
40
25
Range
20
Theoretical Range
15
10
Distance
Distance
30
Range
30
Theoretical Range
20
10
5
0
0
40
60
80
Pressure
100
40
60
80
Pressure
Compare rocket 1 to rocket 2 in range
and theoretical range at 500 angle
Rocket 1= .073 kg and Rocket 2 = .086 kg
100
0 sin
0
0  − R
Students Real Experiment
Example 1:
An air compressor system projectile is fired at 45 0 with initial
velocity 14.16 m/sec. it took 3.4 seconds to reach the ground.
Investigate the maximum range in theory and compare it to the
reality.
2
x 
v 0
cos

x  14 . 16 cos 45
0
 t ,
y 
v 0
sin
 t

1
gt
2
2
1
0
2
( 3 . 4 ), y  14 . 16 (sin 45 )(1 . 7 )  ( 9 . 8 )(1 . 7 )
2

x =14.16(0.7)(3.4)
,
X= 34.04 m ( the range )
Y= 14.16(0.7)(1.7)- 0.5(9.8)(1.72)
y= 2.86 m the maximum height
In the case of y, we substitute for1 one half of the time period
to find the maximum height. 2 t
9
Students measured the range and it was 16.10 meters
which is different totally than their computation
34.04 meter.
Students were looking for explaining to the
resulting error which was -17.94
They came up with Air resistant as a
between their math work and the
They set up this equation for horizontal velocities that
oppose wind speed.
v x  v 0 cos   v R
Such as vR is the wind velocity
By taking the derivative of both sides they will get
the range formula
x
dv
dt
 v 0 (cos  ) t  v R t
wind speed and direction at Paterson,
NJ
http://www.friendlyforecast.com/usa/
archive/archive.php?region=NJ&id=204
500&?-Forecast-Paterson-NewJersey&date=20130524000000&sort=h
our
various among 8.1 mph and 6.9mph
and 4.6 m/h between time 12PM
to 4:00 PM .
The students estimated the average
which is 7 m/h or 3.1 m/s NW
We were launching projectile
towards North so students decided
to multiply speed of Air by Cos (450)
The resulting range is
X= (14.16 (cos 45o)(3.4))- ((5 cos 450)(3.4))
= 22.02 m it is closer to the
reality but with error of -5.92
so our result is still not
accurate Why ?
Physics turn is ended , Calculus starts to tell us
the fact of having errors in our work.
Now , we need to search about the affect of
drag coefficient and Air Resistant.
Assume that our projectile is launched at t =0
making an angle ө to the horizontal.
In addition to the force of gravity, the projectile is subject to an
air resistance, We get some idea of how air resistance and drag
force modify projectile trajectories.
The equation of projectile`s motion is F = mg – cv
Such as c is constant of drag and resistant force, it
depends on the size of projectile and its mass.
a is gravity acceleration 9.8 m/s , v is the velocity , m is
the rocket mass.
ma = mg- cv , then
m
dv
dt
 mg  cv
Using differential equation to separate parts:
dv x
 g
dt
vx
vt
and
vy 


  g  1 

dt
v
t 

dv y
Students have learned that terminal velocity vt
vt 
2w
c d pA
Is affected by the chemistry of the
atmosphere and the gravitational
constant of the earth or any planet . It
is velocity that rocket can not
accelerate after it.
Terminal velocity - the constant maximum velocity
reached by a body falling through the atmosphere under
the attraction of gravity
 v

x
ln
v
 x0
 g

t

vt

And take e for both sides
 gt
v x  v 0 cos  e
x
v 0 v t cos 
g
vt
1  e
Another integration to both sides
 gt / v t

Stop for a moment and go back to physics
Calculate the terminal velocity
W is weigh , cd is drag
coefficient , P is pressure
A is cross sectional area of
projectile
vt 
2w
c d pA
vt 
m
2 mg
c d pA
Is mass of our rocket 7.3gram = .0073 Kg
P is the density of air (p = 1.29 kg/m^3)
A is cross sectional area  =
=.01272  , Where r= .5 inch = .0127 meter
Cd is drag coefficient depends on the mass
Let is calculate our projectile `s drag coefficient.
Net force = Drag – Weight = 0
F

1
2
c d pAv
2
 mg  0
Solve it for cd is from 4 and
1.13 it is calculated by Mr.
Oettinger science teacher
Substitute for all values
vt 
2 mg
c d pA
2  . 0073  9 . 8
vt 
1 . 13  1 . 29  . 0127 
Vt = 13.9 m/s2
2

Back again to maximum range :
x
v 0 v t cos 
g
1  e
 gt / v t

Substitute for all values
x
(14 . 16 )(13 . 92 ) cos 45
9 .8
0
1  e
 ( 9 . 8 )( 3 . 4 ) / 13 . 92
X= 12.92 meter which is matched to the actual
experiment reading with error of 3 m

Go to web-site projectile
motion animation:
projectile-motion/projectilemotion_en.html
A baseball is hit from
2.8 feet above the
ground with an initial
velocity of 160 ft/sec
o
at an angle of 25 from
the horizontal. A gust
component of -9 ft/sec
in the horizontal
direction to the initial
velocity.
25
The Parametric Equations become:
1
X= ( V0cos Ө-VR)t , y = y0+(v0sinӨ)t-
gt
2
2
Given: v = 160ft/s , y = 2.8 ft, g = 32 ft/s
0
1
2
0
g
2
2
 16 ft / s
2
  25
ِِِWe could find the time by using
the formula:
t 
0
v 0 sin 
 t
g
160 sin 25
0
 2 . 1 sec
32
The time to reach the ground is 2t = 4.2 sec
26
The Range:
X= (160cos(25) -9)4.2 = 285.6 ft
The Maximum Height (Altitude):

y= 2.8+(160sin(25 )(2.1))-(0.5(32)(2.1) 2 )
y= 74.2 ft
27
These equations can be modeled graphically by using
Ti- 84 to model the path of the ball:
First set your calculator on Parametric Version and
Degree System.
28
window:
29
Graph the equations:
30
Press trace to find out the maximum height
Is 74.2 ft will be at the time 2.1 second.
31
And the Range is 584.8 ft at time 4.3
seconds or 4.2 seconds.
32
Example 3
Ball tossed with 10 m/s
vertical velocity from
window 20 m above
ground.
Determine:
• velocity and elevation
above ground at time t,
• highest elevation reached
by ball and corresponding
time, and
• time when ball will hit the
ground and corresponding
velocity.
33
SOLUTION:
• Integrate twice to find v(t) and y(t).
dv
 a   9 . 81 m s
dt
v t 
t
 dv    9 . 81 dt
v0
2
v t   v 0   9 . 81 t
0
m

v t   10
  9 . 81
t
2
s 
s 
m
dy
 v  10  9 . 81 t
dt
y t 
t
 dy   10  9 . 81 t dt
y0
y t   y 0  10 t  1 9 . 81 t
2
0
m 2

 m
y t   20 m   10
t

4
.
905

t

2
s 


s 
34
2
• Solve for t at which velocity equals zero
and evaluate corresponding altitude.
m

v t   10
  9 . 81  t  0 t  1 . 019 s
2
s 
s 
m
m
m  2


y t   20 m   10
t

4
.
905

t

2
s 


s 
m
m 


2
y  20 m   10
 1 . 019 s    4 . 905 2  1 . 019 s 
s 


s 
y  25 . 1 m
35
• Solve for t at which altitude equals zero
and evaluate corresponding velocity.
m 2
 m 
y t   20 m   10  t   4 . 905
t  0
2
t   1 . 243 s  meaningles
s  

s 
t  3 . 28 s
s
m 

v  t   10
  9 . 81
t
2
s 
s 
m
m 

v  3 . 28 s   10
  9 . 81
  3 . 28 s 
2
s 
s 
m
v   22 . 2
m
s
36
• You wish to fire an anti-aircraft shell
to intercept an enemy plane flying
towards you at 600 mph and an
altitude of 42,000ft. If the plane is
initially 20 miles away, and your
artillery piece has a muzzle velocity
of 2000 ft/sec, how long do you
have to adjust the piece to an angle
of 600 and fire?
37
38
y
a
vo
60
x
39
y0
=0
V0y , = vo sin 60 =
ay = - 32 ft/s2 .
(.866)(2000) = 1800 ft/sec.
The general equations of motion for constant acceleration in 2-dimensions are:
x (t ) 
1
y (t ) 
1
a t
2 x
a
2
v
v
x
y
y
t
2
2
 v
0x
 v
0
t  x
0
yt  y
0
(t )  a t  v 0 x
x
(t )  a
y
t  v0 y
40
We insert the known values for acceleration &
initial conditions and obtain the specific equations
for the2 shell: x(t) = 1000 t ;
y(t) = - (1/2)(32)t + 1800 t vy(t) =- 32 t + 1800 .
We can now answer any question regarding the
motion of the shell. One question that might occur
to us is: Can we hit a plane at 42,000 ft.? Let's
determine the maximum height to which the shell
will rise.
41
vy(t') = 0 = -32 t' + 1800 , t' = 56.3 sec.
2
Then: ymax = y(t') = -(1/2)(32)(56.3) + 1800 (56.3) =
50,625 ft.
We can reach the elevation of the plane. The next
question is: When will the shell be at the plane's
elevation (42,000 ft)?
2
t
2
y(t') = 42,000 = - 16 t2 + 1800 t
we could solve it for y by using graphing calculator.
42
43
44
By using the graphing calculator we have
42,000 = - 16 t2 + 1800 t
.
The graphing process yields the results:
t1 = 33.03 sec; t2 = 79.47 sec.
45
At the first time the shell will be a
distance x (t'=33.03) = 33,030 ft (6.256
mi) from where it was fired. At the
second time x (t'= 79.47) = 79,470 ft
(15.05 mi).
But the plane traveling at 880 ft/sec will
cover a distance of 13.245 mi in 79.47
seconds. Hence, we can't hit the plane
with the shell on the way down.
46
We illustrate in the figure the various distances
involved. xh = 6.256 mi is position where we
expect to hit the plane; xo = 20 mi is initial
position; xf = ?? is position of plane when we
fire the shell. Since the plane will travel a
distance of (880)(33.03)/(5280) = 5.505 mi
while the shell is in the air, then xf = 6.256 +
5.505 = 11.76 mi away.
47
y
vo
xh
xf
x
o
48
Now the question is, "How long do we wait till
the plane reaches xf? Taking the difference (20 11.76) and dividing by the planes speed, gives
us 49.44 seconds. There is one other 'minor'
question we should ask ourselves. Where will
the plane drop its bomb??? If it will release its
bomb at a distance greater than the 11.76
miles then we better get the heck out of
here!!! A falling bomb will have equations of
motion given by: (using CS with y=0 the
ground).
49
xB(t) = 880 t yB(t) = -(1/2)(32) t2 +42,000 .
Hence, the time to 'fall" is:
yB(t') = 0 =
-16t2 + 42,000  t' = 51.23 sec.
In this time, the bomb travels a horizontal
distance of (880)(51.23)/5280 = 8.54 mi. Thus
the plane plans to release its bomb when it is
8.54 miles from us. But we know the plane
will never reach this point. It will be shot
down 11.76 miles away!
50
Particle and projectile in
Calculus test:
Try : hand out
sheet:
Summary:
We have presented the following:
1) Real Class Experiment to study the affect of air resistance and drag force
on the accuracy of projectile.
2) The perfect elevation angle of launching rocket with and without air resistance.
3) Use differential equation to compute terminal velocity , drag coefficient
and terminal velocity.
4) Baseball motion and time to reach the ground under the condition of
baseball playground. (sport application).
5) Ball tossed with vertical velocity from tower window.
6) Firing an anti-aircraft shell to intercept an enemy plane flying towards
you(Military application).
7) AP- Calculus sample problems.
Citation:
http://www.friendlyforecast.com/usa/archive/archive.php?region=NJ&id=204500&?Forecast-Paterson-New-Jersey&date=20130524000000&sort=hour
http://www.grc.nasa.gov/WWW/k-12/rocket/termvr.html
http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node29.html
http://demonstrations.wolfram.com/ProjectileWithAirDrag/
http://www.rocketmime.com/rockets/rckt_eqn.html
Books resources:
AP Calculus ninth edition for: Ron Larson, The Pennsylvania State University.
Bruce H. Edwards, University of Florida
The physics Book for Clifford A. Pickovee