Transcript Vectors: Motion and Forces in Two Dimensions

do now
A
+
B
=
?
• The wrong diagrams
Draw the right diagram for A + B
3.3 projectile motion
Objectives
1. Recognize examples of projectile motion.
2. Describe the path of a projectile as a
parabola.
3. Resolve vectors into their components
and apply the kinematic equations to solve
problems involving projectile motion.
question
• The long jumper builds up speed in the xdirection and jumps, so there is also a
component of speed in the y direction. Does the
angle of take-off matter to the jumper?
• A parabola is the set of all
points in the plane
equidistant from a given line
(the conic section directrix)
and a given point not on the
line (the focus).
y ~ x2
What is a projectile?
• An object that is launched into the air
with some INITIAL VELOCITY
• Can be launched at ANY ANGLE
• In FREEFALL after launch (no outside
forces except force of gravity)
• The path of the projectile is a
PARABOLA
Effect of air resistance
demo
• The vertical
motion of a
projectile is same
as free falling
object, with
constant
acceleration:
• a = -9.81 m/s/s
Two types of projectiles
• Projectile launched horizontally
• Projectile launched at an angle
Projectile Launched horizontally
Projectile Launched at an angle
Over the Edge
Horizontal Projectiles
Projectiles Launched Horizontally
Vertical motion:
• Free fall free rest:
vy = 0
ay = -9.81 m/s2
Horizontal motion:
• vx = vix = constant
• ax = 0
• Thus, the projectile travels with a
constant horizontal velocity and a
downward vertical acceleration.
Velocity of Horizontal Projectiles
• Horizontal motion is
constant: velocity is
constant.
• Vertical: same as drop
the ball from rest:
velocity is increasing by
9.81 m/s every second
Projectile launched horizontally
•
•
•
•
•
A projectile is any object upon which the only force is
gravity
Projectiles travel with a parabolic trajectory due to the
influence of gravity,
In horizontal direction, the projectile has no
acceleration, its velocity is constant: vx = vi
In vertical direction, the projectile has acceleration:
a = -9.81 m/s/s. Its initial vertical velocity is zero and it
changes by -9.81 m/s each second. Same as an object
falling from rest.
The horizontal motion of a projectile is
independent of its vertical motion
Example
•
An object was projected horizontally from a tall cliff.
The diagram represents the path of the object,
neglecting friction.
• Comparing the following at point A & B:
1. Acceleration
2. Horizontal velocity
3. Vertical velocity
As the red ball rolls off the edge, a green
ball is dropped from rest from the same
height at the same time
Which one will hit the ground first?
They will hit
at the SAME
TIME!!!
The same time?!? How?!?
The red ball has an initial
HORIZONTAL velocity (vix)
But does not have any initial
VERTICAL velocity (viy = 0)
vix
The green ball falls from rest
and has no initial
velocity IN EITHER
DIRECTION!
viy and vix = 0
One Dimension at a Time
• Both balls begin with no VERTICAL
VELOCITY
• Both fall the same DISTANCE
VERTICALLY
• Find time of flight by solving in the
appropriate dimension
We can find an object’s displacement in
EITHER DIMENSION using TIME
Example #1
• A bullet is fired horizontally from a gun that is
1.7 meters above the ground with a velocity of
55 meters per second.
• At the same time that the bullet is fired, the
shooter drops an identical bullet from the same
height.
– Which
bullet
hits the
Both hit
the ground
atground
the samefirst?
time
Equation for horizontal projectile
Horizontal
• ax = 0
θ=0
d = ½ (vi + vf)t
vix = vicosθ = vi
x = vix∙t
Vertical
• ay = -9.81 m/s2
vf = vi + at
viy = visinθ = 0
y = ½ (viy + vfy)t
d = vit + ½at2
vfy = viy + ayt
vf2 = vi2 + 2ad
y = viyt + ½ayt2
vfy2 = viy2 + 2ayy
x and y has the same t
Example #2
• An airplane making a supply drop to
troops behind enemy lines is flying with
a speed of 300 meters per second at an
altitude of 300 meters.
– How far from the drop zone should the
aircraft drop the supplies?
Need time from vertical
Use time in horizontal
dy = viyt + ½ ayt2
300 m = 0 + ½ (-9.81 m/s2)t2
t = 7.82 s
dx = vixt + ½ axt2
dx = (300 m/s)(7.82 s) + 0
dx = 2346 m
Example #3
Need time from vertical
dy = viyt + ½ ayt2
45 m = 0 + ½ (-9.81 m/s2)t2
t = 3.03 s
Use time to find v
v=d/t
v = 15 m / 3.03 s
v = 4.95 m /s
• A stuntman jumps off
the edge of a 45 meter
tall building to an air
mattress that has been
placed on the street
below at 15 meters from
the edge of the building.
– What minimum initial
velocity does he need in
order to make it onto the
air mattress?
Example
#4
Example #4
• A CSI detective investigating an accident
scene finds a car that has flown off the
edge of a cliff. The car is 79 meters from
the edge of the 25 meter high cliff.
– What was the car’s initial horizontal velocity
as it went off the edge?
Need time from vertical
Use time in horizontal
dy = viyt + ½ ayt2
25 m = 0 + ½ (-9.81 m/s2)t2
t = 2.26 s
dx = vixt + ½ axt2
79 m = vix (2.26 s) + 0
vix = 34.96 m/s
Example #5
• The path of a stunt car driven horizontally off a cliff is
represented in the diagram below. After leaving the cliff, the
car falls freely to point A in 0.50 second and to point B in
1.00 second. 1. Determine the magnitude of the horizontal
component of the velocity of the car at
point B. [Neglect friction.]
2. Determine the magnitude of
the vertical velocity of the car
at point A.
3. Calculate the magnitude of
the vertical displacement, dy,
of the car from point A to
point B. [Neglect friction.]
Class work
• Page 101 - Sample problem 3D
• Page 102 – practice 3D
Answers:
1.0.66 m/s
2.4.9 m/s
3.7.6 m/s
4.5.6 m
Do now
Fire Away!!!
Projectiles
Launched at an Angle
Projectile Launched at an angle
Projectile Vector Diagram
t1/2
vfy = 0 (at top)
ttot
Initial velocity
What
What
Pythagorean
isisthe
thehorizontal
vertical
Theorem
part
partofof
2 = v 2initial
2 velocity?
the soccerviball’s
+
v
ix
iy
viy
cos θθ
ix = vi sin
12 m/s
6 m/s
30°
10.4 m/s
What do we know or assume about the
vertical part of a projectile problem?
• Initial vertical velocity = vi sin θ
• Acceleration = -9.81 m/s2
• Vertical speed will be 0 at the maximum height
• Time to top = HALF total time in the air
– Find time to top using final velocity equation
• Vertical Distance – Max height
– Use time to top and solve vertical distance equation
What do we know or assume about the
horizontal part of a projectile problem?
• Initial vertical velocity = vi cos θ
• Acceleration = 0 (if we assume no air resistance)
• Horizontal Distance – Range
– Use total time and solve horizontal distance equation
The symmetrical nature of a ground
launched projectile
•What is the acceleration at the top of the path?
•What is the vertical velocity at the top of the path?
Projectile launched at an angle
•
•
•
•
•
A projectile is any object upon which the only force is
gravity
Projectiles travel with a parabolic trajectory due to the
influence of gravity,
In horizontal direction, the projectile has no
acceleration, its velocity is constant: vx = vicosθ
In vertical direction, the projectile has acceleration:
a = -9.81 m/s/s. Its velocity of a projectile changes by
-9.81 m/s each second. Same as a free falling object.
The horizontal motion of a projectile is
independent of its vertical motion
• A projectile is fired with initial horizontal velocity at 10.00 m/s,
and vertical velocity at +20. m/s. Determine the horizontal and
vertical velocity at 1 – 5 seconds after the projectile is fired.
Use g = 10 m/s/s.
Time
0s
1s
2s
3s
4s
5s
Horizontal
Velocity
Vertical
Velocity
Accelartion
Equation for projectile motion
1D equation:
d = ½ (vi + vf)t
vf = vi + at
d = vit + ½at2
vf2 = vi2 + 2ad
Horizontal
• ax = 0
Vertical
• ay = -9.81 m/s2
viy = visinθ
vfy = 0 (at top)
vfy = - viy (at same height)
y = ½ (viy + vfy)t
vfy = viy + ayt
vix = vicosθ
y = viyt + ½ayt2
x = vix∙t
vfy2 = viy2 + 2ayy
Initial Velocity Components
• Since velocity is a vector quantity, vector
resolution is used to determine the components
of velocity.
v2 = v 2 + v 2
i
ix
iy
SOH CAH TOA
viy
vi
θ
vix
sinθ = viy / vi
viy = visinθ
cosθ = vix / vi
vix = vicosθ
• Special case: horizontally launched projectile:
• θ = 0o: viy = visinθ = 0; vix = vicosθ = vi
Examples Determine the horizontal
and vertical components
1. A water balloon is launched with a speed of
40 m/s at an angle of 60 degrees to the
horizontal.
2. A motorcycle stunt person traveling 70 mi/hr
jumps off a ramp parallel to the horizontal.
3. A springboard diver jumps with a velocity
of 10 m/s at an angle of 80 degrees to the
horizontal.
example
• A machine fired several projectiles
at the same angle, θ, above the
horizontal. Each projectile was
fired with a different initial
velocity, vi. The graph below
represents the relationship
between the magnitude of the
initial vertical velocity, viy, and the
magnitude of the corresponding
initial velocity, vi, of these
projectiles. Calculate the
magnitude of the initial horizontal
velocity of the projectile, vix, when
the magnitude of its initial
velocity, vi, was 40. meters per
second.
• The point of resolving an initial velocity
vector into its two components is to use
the values of these two components to
analyze a projectile's motion and
determine such parameters as
– the horizontal displacement,
– the vertical displacement,
– the final vertical velocity,
– the time to reach the peak of the trajectory,
– the time to fall to the ground, etc.
Example
• A football is kicked with an initial velocity of 25 m/s at an
angle of 45-degrees with the horizontal. Determine the
time of flight, the horizontal displacement, and the peak
height of the football.
Horizontal Component
Vertical Component
ax = 0 m/s/s
vix = 25 m/s•cos45o= 17.7 m/s
x=?
ay = -9.81 m/s/s
viy=25 m/s•sin45o= 17.7 m/s
vfy = -17.7 m/s
ymax = ?
ttotal = ?
• Solve for t – use vertical information:
vfy = viy + ayt
t = 3.61 s
• Solve for x – use horizontal information:
x = vix•t + ½ •ax•t2
x = 63.8 m
• Solve for peak height – use vertical information:
vfy2 = viy2 + 2ayypeak
At the top, vfy = 0
ypeak = 15.9 m
example
• A long jumper leaves the ground with an initial velocity of
12 m/s at an angle of 28-degrees above the horizontal.
Determine the time of flight, the horizontal distance, and
the peak height of the long-jumper.
Horizontal Component
Vertical Component
ax = 0 m/s/s
vix = vi•cosθ
vix = 12 m/s•cos28o
vix = 10.6 m/s
ttotal = 2tup
x=?
ay = -9.8 m/s/s
viy = vi•sinθ
viy = 12 m/s•sin28o
viy = 5.6 m/s
vfy = -5.6 m/s
y=?
tup = ?
t = 1.1 s
x = 12.2 m
ypeak = 1.6 m
Example
•
1.
2.
A cannon elevated at an angle of 35° to the
horizontal fires a cannonball, which travels the path
shown in the diagram. [Neglect air resistance and
assume the ball lands at the same height above the
ground from which it was launched.] If the ball lands
7.0 × 102 meters from the cannon 7.0 seconds after
it was fired,
what is the horizontal component of its initial
velocity?
what is the vertical component of its initial velocity?
3. An object is thrown horizontally off a cliff
with an initial velocity of 5.0 meters per
second. The object strikes the ground
3.0 seconds later. What is the vertical
speed of the object as it reaches the
ground? [Neglect friction.]
4. In the diagram, a 10.-kilogram sphere, A, is projected
horizontally with a velocity of 30. meters per second
due east from a height of 20. meters above level
ground. At the same instant, a 20.-kilogram sphere, B ,
is projected horizontally with a velocity of 10. meters
per second due west from a height of 80. meters above
level ground. [Neglect air friction.] Initially, the spheres
are separated by a horizontal distance of 100. meters.
What is the horizontal separation of the spheres at the
end of 1.5 seconds?
Total flight time, range, max height
Time to go up
• t = visinθ / g
• As θ increases, flight time
increase.
Max time: θ = 90o
Range
• Range = vi2sin2θ /g
• Projectile has maximum
range when θ = 45o
Max height
• hmax = (visinθ)2/2g
• As θ increases, flight
height increase.
Max height: θ = 90o
Class work
• Page 103 – sample problem 3E
• Page 104 – practice 3E #1-5
Answers:
1. Yes
2. 70.3 m
3. 20. s; 4.8 m
4. 6.2 m/s
5. 17.7 m/s; 6.60 m