Transcript Ideal Projectile Motion

Ideal Projectile
Motion
SECTION 10.4A
Ideal Projectile Motion
What do we mean by the word “ideal” here?
Assumptions:
• The projectile behaves like a particle moving in a
vertical coordinate plane.
• The only force acting on the projectile during its
flight is the constant force of gravity.
• The projectile will follow a trajectory that is
perfectly parabolic.
Ideal Projectile Motion
Position, velocity, acceleration due to gravity, and launch angle
at t = 0…
Initial velocity:
v 0   v 0 cos   i   v 0 sin   j
  v0 cos   i   v0 sin   j
v0
Initial position:
v 0 sin  j

r = 0 at
time t = 0
v 0 cos  i
a   gj
r0  0i  0 j  0
Ideal Projectile Motion
Position, velocity, and acceleration at a later time t…
 x, y 
r  xi  yj
v
a   gj
R
Horizontal range
Ideal Projectile Motion
Newton’s second law of motion: the force acting on the
projectile is equal to the projectile’s mass times its acceleration.
For ideal motion, this force is solely the gravitational force:
2
d r
m 2  mgj
dt
Let’s solve this initial value problem,
using initial conditions: r  r and
2
d r
  gj
2
dt
dr
 v 0 when t  0
0
dt
dr
First integration:
   gt  j  v 0
dt
1 2
Second integration: r   gt j  v 0t  r0
2
Ideal Projectile Motion
1 2
r   gt j  v 0t  r0
2
Previous equations:
v 0   v0 cos   i   v0 sin   j
r0  0i  0j  0
Substitution:
1 2
r   gt j   v0 cos   ti   v0 sin   tj  0
2
1 2

  v0 cos   ti    v0 sin   t  gt  j
2


Ideal Projectile Motion
1 2

r   v0 cos   ti    v0 sin   t  gt  j
2


This is the vector equation for ideal projectile motion. The
angle  is the projectile’s launch angle (firing angle,
angle of elevation), and v0 is the projectile’s initial speed.
Break into a pair of scalar equations:
x   v0 cos   t
1 2
y   v0 sin   t  gt
2
These are the parametric equations for ideal projectile
motion. Gravity constants:
m
ft
g  9.8 2  32 2
sec
sec
Ideal Projectile Motion
1 2

r   v0 cos   ti    v0 sin   t  gt  j
2


1 2
y   v0 sin   t  gt
x   v0 cos   t
2
If the ideal projectile is fired from the point
of the origin:
x  x0   v0 cos   t
 x0 , y0  instead
1 2
y  y0   v0 sin   t  gt
2
Practice Problem
A projectile is fired from the origin over horizontal ground at
an initial speed of 500 m/sec and a launch angle of 60 degrees.
Where will the projectile be 10 sec later?
Parametric equations for this situation, evaluated at this time:
x   500 cos 60 10   2500
1
2
y   500sin 60 10    9.8 10   3840.127
2
Ten seconds after firing, the projectile is about
3840.127 m in the air and 2500 m downrange
Height, Flight Time, and Range
The projectile reaches its highest point when its vertical velocity
component is zero:
1
y   v0 sin   t  gt
2
v0 sin 
dy
 v0 sin   gt  0
t
dt
g
2
Height at this time:
 v0 sin   1  v0 sin  
y   v0 sin   
 g

 g  2  g 
v0 sin  


g
2
2
g  v0 sin  
v0 sin  



2
2g
2g
2
2
Height, Flight Time, and Range
To find the total flight time of the projectile, find the time it
takes for the vertical position to equal zero:
1 2
y   v0 sin   t  gt  0
2
1 

t  v0 sin   gt   0
2 

2v0 sin 
t  0, t 
g
Height, Flight Time, and Range
To find the projectile’s range R, the distance from the origin to
the point of impact on horizontal ground, find the value of x
when t equals the flight time:
2v0 sin 
t
x   v0 cos   t
g
 2v0 sin   v0
R   v0 cos   
 2sin  cos  

g

 g
2
2
v0

sin 2
g
Note: The range is largest when sin 2  1 or   45.
Height, Flight Time, and Range
For ideal projectile motion when an object is launched from
the origin over a horizontal surface with initial speed v0 and
launch angle  :
Maximum Height:
Flight Time:
2v0 sin 
t
g
2
Range:
ymax
v0 sin  


v0
R
sin 2
g
2g
2
Practice Problems
Find the maximum height, flight time, and range of a projectile
fired from the origin over horizontal ground at an initial speed
of 500 m/sec and a launch angle of 60 degrees. Then graph the
path of the projectile.
2
Max Height:
Flight Time:
500sin 60 

ymax 
 9566.327 m
2  9.8 
2  500  sin 60
 88.370 sec
t
9.8
Range:
500 

R
9.8
2
sin120  22, 092.485
m
Practice Problems
Find the maximum height, flight time, and range of a projectile
fired from the origin over horizontal ground at an initial speed
of 500 m/sec and a launch angle of 60 degrees. Then graph the
path of the projectile.
Parametric equations for the path of the projectile:
x   500 cos 60  t  250t


1
2
y   500sin 60  t   9.8  t  250 3 t  4.9t 2
2
Graph window: [0, 25000] by [–2000, 10000]
Practice Problems
To open the 1992 Summer Olympics in Barcelona, bronze
medalist archer Antonio Rebollo lit the Olympic torch with a
flaming arrow. Suppose that Rebollo shot the arrow at a height
of 6 ft above ground level 30 yd from the 70-ft-high cauldron,
and he wanted the arrow to reach a maximum height exactly
4 ft above the center of the cauldron.
y
v0
6
ymax  74

90
x
Practice Problems
(a) Express
angle .

ymax
in terms of the initial speed
ymax
v0 sin  


2g
v0
and firing
2
6
(b) Use ymax  74 ft and the result from part (a) to find the
value of v sin  .
v0 sin  

74 
2  32 
0
 v0 sin  
64
2
6
2
 68
v0 sin   4352
Practice Problems
(c) Find the value of v0 cos  .
x  x0   v0 cos   t
Values when the arrow reaches maximum height:
 v0 sin  
90  0   v0 cos   

 g 
 4352 
90   v0 cos   

 32 
2880
v0 cos  
4352
Practice Problems
(d) Find the initial firing angle of the arrow.
v0 sin   4352
2880
v0 cos  
4352
Put these parts together:
4352
v0 sin 
4352 68
 tan  


2880 4352 2880 45
v0 cos 
 68   56.505
  tan  
 45 
1