physics. good news/bad news: these are the same formulas we used for linear motion. do you know...
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Physics
Good News/Bad News: These are the same formulas we used for linear motion. Do you know them?
22
1 attvd i atvv if
advv if 222 If the answer is “NO”, then get familiar with them NOW!
We assume NO AIR RESISTANCE! (Welcome to “Newtonia”), therefore… The path of a projectile is a parabola. Horizontal motion is constant velocity.
Vertical motion is in “free-fall”.
Vertical velocity at the top of the path is zero
Time is the same for both horizontal and vertical motions.
0
constant
x
x
a
v
2m/s 81.9ya
0top
yv
horizontal or “x” – direction
xxixfx
xixfx
xixx
davv
tavv
tatvd
222
22
1
vertical or “y” – direction
yyiyfy
yiyfy
yiyy
davv
tavv
tatvd
222
22
1
Remember that for projectiles, the horizontal and vertical motions must be separated and analyzed independently. Remember that “ax” is zero and “ay” is acceleration due to gravity “g = 9.81 m/s2”.
0
0
0
A cannon ball is shot horizontally from a cliff.
vx
dy
Range, dx
What do we know? For all projectiles…
0
constant
x
x
a
v
2m/s 81.9ya
yx
yiy
ttt
vv
0top
Hint: You should always list your known values at the beginning of any problem and assign those values variables.
A cannon ball is shot horizontally from a cliff.
vx
dy
Range, dx
0
constant
x
x
a
v
2m/s 81.9ya
yx
yiy
ttt
vv
0top
Remember to keep the horizontal and vertical motions separate. Time is the factor that ties them together. Since the vertical motion is treated like “free-fall” and we have more info about the vertical, we should use that to find time first.
Knowns:
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
dy=35 m
Range, dx
0
constant
x
x
a
v
2m/s 81.9ya
yx
yiy
ttt
vv
0top
Add the given values to our list of known values. Now that the diagram is drawn and labeled and we have identified and listed all of our “known” and “given” values for the problem, let’s begin by finding time.
Knowns:Givens:
m35
m/s 5
d
v
y
x
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
dy=35 m
Range, dx
0
constant
x
x
a
v
2m/s 81.9ya
yx
yiy
ttt
vv
0top
Since we know more values for vertical motion, let’s use it to find time. Start with the distance equation…
Knowns:Givens:
m35
m/s 5
d
v
y
x
22
1
22
1
)81.9(035 t
attvd yiy
Now solve for t… sec67.2905.4
35
905.435 2
t
t
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
dy=35 m
Range, dx
0
constant
x
x
a
v
2m/s 81.9ya
yx
yiy
ttt
vv
0top
Now that we know time, let’s find dx. Remember that horizontal motion is constant. Let’s use the distance formula again. This time in the x – direction…
Knowns:Givens:
m35
m/s 5
d
v
y
x
st 67.2Calculated:
m 36.13s 67.2m/s 5
22
1
x
xxix
d
tatvd 0
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
dy=35 m
Range, dx
0
constant
x
x
a
v
2m/s 81.9ya
yx
yiy
ttt
vv
0top
Final velocity requires a little more thought. Remember that velocity is a vector quantity so we must state our answer as a magnitude (speed that the projectile strikes the ground) and direction (angle the projectile strikes the ground). Also remember horizontal velocity is constant, therefore the projectile will never strike the ground exactly at 90°. That means we need to look at the horizontal (x) and vertical (y) components that make up the final velocity.
Knowns:Givens:
m35
m/s 5
d
v
y
x
st 67.2Calculated:
m 36.13xd
Vf
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
dy=35 m
Range, dx
0
constant
x
x
a
v
2m/s 81.9ya
yx
yiy
ttt
vv
0top
Let’s look more closely at the vector, vf. To help see it better, let’s exaggerate the angle. Since x- and y- motion are separate, there must be components.
Knowns:Givens:
m35
m/s 5
d
v
y
x
st 67.2Calculated:
m 36.13xd
Vf
Vf
Vfx = 5 m/s
Vfy
θ
θ
So, we have the x-component already due to the fact that horizontal velocity is constant. Before we can find vf, we must find the vertical component, vfy.
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
dy=35 m
Range, dx
0
constant
x
x
a
v
2m/s 81.9ya
yx
yiy
ttt
vv
0top
To find vfy, remember that vertical motion is in “free-fall” so it is accelerated by gravity from zero to some value just before it hits the ground.
Knowns:Givens:
m35
m/s 5
d
v
y
x
st 67.2Calculated:
m 36.13xd
Vf
Vf
Vfx = 5 m/s
Vfy
θ
θ
atvv yiyf
m/s 2.26
)67.2)(m/s 81.9(0 2
yf
yf
yiyf
v
sv
atvv
Calculating vfy:
Still not finished. Gotta put components together for final.
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
dy=35 m
Range, dx
0
constant
x
x
a
v
2m/s 81.9ya
yx
yiy
ttt
vv
0top
Now we know both the x- and y- components of the final velocity vector. We need to put them together for magnitude and direction of final velocity.
Knowns:Givens:
m35
m/s 5
d
v
y
x
st 67.2Calculated:
m 36.13xd
Vf
Vf
Vfx = 5 m/s
θ
θ
atvv yiyf
direction 2.79tan
(speed) magnitude m/s 7.26
2.265
52.261
2222
f
yfxf
v
vvv
Putting it together to calculate vf:
Final Velocity = 26.7 m/s, 79.2°
vi
θ
Since the initial velocity represents motion in both the horizontal (x) and vertical (y) directions at the same time, we cannot use it in any of our equations. Remember, the most important thing about projectiles is that we must treat the horizontal (x) and vertical (y) completely separate from each other. So…we need to separate “vi” into its x- and y- components. We will use the method we used for vectors.
viy
θ
vix
vix
vi
sin
cos
iiy
iix
vv
vv
Now that we have the components
of the initial velocity, we will use only those for calculations. **Never use the original velocity at the angle in an equation!
viy
viθ
A projectile’s path is a parabola, ALWAYS. That means if a projectile is launched and lands at the same height, there will be symmetry. The angle of launch and angle of landing will be equal. The initial velocity and the final velocity will be the same magnitude. Also, that means the components will be the same.
x
fyv
v
fyxf vvv
1
22
tan
viy
vix
vfθ
vfx
vfy
Since the horizontal motion is always at constant velocity…
xxixf vvv
Since the vertical motion is the same as a ball that is thrown straight up or dropped straight down (in free-fall), the y-components are equal and opposite.
yiyf vv
vfθ
vfx
vfy
vi
θStep 1: List known values! Draw and label picture.
Knowns (for all projectiles):
0
constant
x
x
a
v
2m/s 81.9ya
yx
y
ttt
v
0top
Vy top = 0
Viy
Vx
dymax =height
Range, dx
vi
θ
Knowns (for all projectiles):
0
constant
x
x
a
v
2m/s 81.9ya
yx
y
ttt
v
0top
Vx
Vy top = 0
Vy dymax =height
Range, dx
Step 2: Divide initial velocity into horizontal (x) and vertical (y) components.
sin
cos
iiy
iix
vv
vv
Step 3: Find time if possible. Use vertical motion.
Keeping the horizontal and vertical motions separate!
vi
θ
Almost every projectile problem can be solved by starting with the displacement equation to solve for time. In this case…
212 0y iy yd v t a t
Now solve for time. Yes, it is a quadratic equation! This will be the time for the entire flight. NOTE: If you want to find maximum height you will only use half the time. If you want to find range, use the total time.
Finding time – Method 1:Since the initial and final vertical positions are both the same, vertical displacement dy = 0.
Vy top = 0
Vy dymax =height
Range, dx
Note: There are three ways to find time for this problem. You may use any of them you wish.
vi
θ
Vy top = 0
Vy
Vx
dymax =height
Range, dx
Find time – Method 2: Use vertical motion and symmetry. Remember that the y component of initial and final velocities are equal and opposite. So using the formula
plug in the known values from symmetry , vfy=-viy and solve for t.
vf
θvfx
vfy
tavv yyiyf
av
t
tavv
yi
yyiyi
2
Note: usually any vectors acting upward such as initial velocity are considered to be positive, therefore acceleration due to gravity is negative (-9.81) so the time will NOT turn out to be a negative.
vi
θ
Vy top = 0
Vy
Vx
dymax =height
Range, dxFind time – Method 3: Remember for all projectiles, the vertical velocity at the very top of the path is zero. If the projectile is launched and lands at the same height, the top of the path occurs at exactly the half-way point. If we use this as either the initial or final velocity we can calculate ½ the time of flight and then simply double it! Beginning with the same equation,
vf
θvfx
vfy
atvv yiyf
81.92
0
yi
yyi
vt
tav Remember to use the symmetry between initial and final y-velocities!
To work from launch to top of path…
OR…To work from top of path to landing …
81.92
0
0
yi
yyi
yyf
vt
tav
tav
V i= 3
0 m/s
60°Step 1: List known values! Draw and label picture.
Knowns (for all projectiles):
0
constant
x
x
a
v
2m/s 81.9ya
yx
y
ttt
v
0top
Vy top = 0
Viy
Vx
dymax =height
Range, dx
A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
Given values:
60
m/s 03
iv
V i= 3
0 m/s
60°
Step 2: Divide initial velocity into x- & y- components.
Knowns (for all projectiles):
0
constant
x
x
a
v
2m/s 81.9ya
yx
y
ttt
v
0top
Vy top = 0
Viy
Vx
dymax =height
Range, dx
A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
Given values:
60
m/s 03
iv
m/s 2660sin30sin
m/s 1560cos30cos
iiy
iix
vv
vv
We can add these to what we know. WE WILL NOT USE THE 30 m/s again in this problem because it is not purely an x- or y- value.
Finding time – Method 1:Remember that dy = 0 because the projectile is starting and ending at the same level (y-position). So, using the known and given values for this problem and the components we calculated, we can solve for time.
Vy top = 0
v y =
26 m
/s
dymax =height
Range, dx
v i= 3
0 m/s
60°
sec 3.5
905.426
)81.9(260
905.426
2
22
1
22
1
t
tt
tt
tatvd yyiy
A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
vix =15 m/s
vi
θ
Vy top = 0
dymax =height
Range, dx
Find time – Method 2: Remember that the y component of initial and final velocities are equal and opposite. So, to calculate…
vf
θ
sec 3.581.952
)81.9(2626
t
t
tavv yyiyf
A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
v y =
26 m
/sv
fy =-26 m
/s
vx =15 m/s vx =15 m/s
Note: This method gives the same answer as the previous one. Choose the one that make the most sense to you and use it.
vi
θ
Vy top = 0
dymax =height
Range, dxFind time – Method 3: Remember for all projectiles, the vertical velocity at the very top of the path is zero. If the projectile is launched and lands at the same height, the top of the path occurs at exactly the half-way point.
vf
θ
sec 3.52*65.2
sec 65.2
)81.9(260
81.926
2
t
t
tavv
t
yyiyfTo work from launch to top of path…
OR…To work from top of path to landing…
v y =
26 m
/sv
y =-26 m
/sA football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
vx =15 m/s vx =15 m/s
sec 3.52*65.2
sec 65.2
)81.9(026
81.926
2
t
t
tavv
t
yyiyf
vi
θ
Vy top = 0
Vy
Vx
dymax =height
Range, dx
vf
θvfx
vfy
Now that I know time, I can add it to my list of known, given, and calculated values. To review…
Knowns (for all projectiles):
0
constant
x
x
a
v
2m/s 81.9ya
yx
y
ttt
v
0top
Givens:
Calculated values:
60
m/s 03
iv
sec3.5
m/s 26
m/s 15
t
v
v
yi
xi
A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
vi
θ
Vy top = 0
dymax =height
Range, dx
Now that we know time, let’s calculate horizontal distance. Remember that horizontal acceleration is zero.
vf
θv y =
26 m
/sv
y =-26 m
/sA football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
vx =15 m/s vx =15 m/s
m 579s) 3.5)( 15(
22
1
.d
tatvd
sm
x
xxix
0
vi
θ
Vy top = 0
dymax =height
Range, dx
Method 1 to find the maximum height for this problem, remember that because it is launched and lands at the same level, maximum height occurs exactly half-way through the flight. So…USE ½ of the total time of flight.
vf
θv y =
26 m
/sv
y =-26 m
/sA football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
vx =15 m/s vx =15 m/s
m 45.34
))( 81.9())( 26( 22
s 3.5s
m2
12
s 3.5
22
1
2
y
sm
y
yyiy
d
d
tatvd
vi
θ
Vy top = 0
dymax =height
Range, dx
Method 2 for finding the maximum height: Since the final position is at the top, again use ½ t and also use vyf = 0.
vf
θv y =
26 m
/sv
y =-26 m
/sA football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
vx =15 m/s vx =15 m/s
2
26 / 0 5.3
2 2
34.45 m
yi yfy
y
y
v vd t
m s sd
d
vi
θ
Vy top = 0
dymax =height
Range, dx
Method 3 for finding the maximum height: Since the final position is at the top, use vyf = 0.
vf
θv y =
26 m
/sv
y =-26 m
/sA football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.
vx =15 m/s vx =15 m/s
2 2
2 2
2
0 (26 / ) 2( 9.8 / )
34.49 m
yf yi y y
y
y
v v a d
m s m s d
d
022
1 tatvd yyiy
Vertical displacement is not zero. Consider the launch point as the zero height and then vertical displacement, dy, will be a positive number (as long as you continue to assume up is positive. Plug the value for dy into the above equation and solve for time. (Hint: Graph it and find the zeros! It’s easier than the quadratic equation.) The answer must be positive. This will be the time for the flight to that point. Then you can use that time to find the horizontal distance the object traveled to get to that point.
22
1 tatvd xxix
viθ
dy
dx
0NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL!