Transcript PPt
Projectile Motion Horizontally
Projectile Motion
Once
a difficult problem for
cannoneers
If
the vertical and horizontal
components are separated, the problem
becomes simple.
Projectile Motion
– simple as a ball
rolling across the table
Use equations like x = v x t
Horizontal
Projectile Motion
– simple as a free falling
gravity problem
Use equations ∆y = (1/2)gt2
and vf = gt
Vertical
Projectile Motion
NOTE:
Air and other resistance is
being ignored for the time
being.
Projectile Motion
Projectile Motion of a Cannon Ball
Projected Horizontally
Question 1
Consider these diagrams to answer the
following questions:
a. Which shows the initial velocity?
a
b. Which shows the acceleration vector?
b
Question 2
What determines the time for the cannon ball to hit the ground?
Question 2 Solution
The time for the cannon ball to hit the
ground is determined by the height from
which it drops.
Question 3
Supposing a snowmobile is equipped with a flare
launcher which is capable of launching a sphere
vertically (relative to the snowmobile). If the
snowmobile is in motion and launches the flare
and maintains a constant horizontal velocity after
the launch, then where will the flare land (neglect
air resistance)?
a. in front of the snowmobile
b. behind the snowmobile
c. in the snowmobile
Solution for Question 3
C
The flare will land inside the snowmobile.
The flare has the same horizontal velocity
as the snowmobile that it was initially in.
The velocity does not change
Simulation:
Constant Velocity in the Horizontal Direction
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vectors/pap.html
Horizontal & Vertical Velocities
Horizontal & Vertical Displacements
Summary of Projectile Motion
Horizontally
Horizontal Motion Vertical Motion
Forces
(Present?- Yes or No) If
present, what direction
No
Yes
The force of gravity acts
downward
Acceleration
No
“g” is downward at -9.81 m/s2
(Present?- Yes or No)
If present, what direction?)
Velocity
(Constant or Changing)
Yes
Constant
Changing
(by -9.81 m/s each second)
Projectile Motion at Zero Angle
An ugly giant rolls a billiard ball with uniform velocity 30 m/s
across the top of his desk. The ball rolls off the edge of the
desk and lands on the floor 120 m from the edge of the desk.
How high is the table?
Height of Table
Time of fall
vx = x/t
30 m/s = 120 m / t
t = 4.0 s
Height of table
∆y = vit + ½ gt2
∆y = ½ (-9.81 m/s2)(4.0s)2
∆y = -78 m………78m
An ugly giant rolls a billiard ball with uniform velocity 30 m/s
across the top of his desk. The ball rolls off the edge of the
desk and lands on the floor 120 m from the edge of the desk.
At what speed will it hit the ground?
Velocity When It Hits The Ground
Horizontally speed is constant.
30 m/s = vx
Vertical Speed
vf = vi + gt
vf = (-9.81 m/s2 )4.0 s
vf = 39 m/s
Resultant speed = 49 m/s
Characteristics & Assumptions
Planar motion
Air resistance: negligible
Gravity: downward
No horizontal acceleration
Constant vertical acceleration
Parabolic trajectory
Independent horizontal and vertical motions
Question 4
Anna Litical drops a ball from rest from the top of
80.-meter high cliff. How much time will it take for the
ball to reach the ground and at what height will the ball
drop after each second of motion?
Solution
∆y = vit + ½ gt2
-80.m = ½ (-9.81 m/s2)t2
-160m = (-9.81 m/s2) t2
t = 4.0 s
Each sec the displacements are as follows:
∆y = vit + ½ gt2 = ½ (-9.81m/s2)(1sec)2
-4.9m, -20m,-44m,-78m
Question 5
A cannonball is launched horizontally from the top of an
80.-meter high cliff. How much time will it take for the
ball to reach the ground and at what height will the ball be
after each second of travel?
Question 5 Solution
∆y = vit + ½ gt2
-80.m = ½ (-9.81 m/s2)t2
-160m = (-9.81 m/s2) t2
t = 4.0 s
Each sec the displacements are as follows:
∆y = vit + ½ gt2 = ½ (-9.81m/s2)(1sec)2
-4.9m, -20m,-44m,-78m
Question 6
A pool ball leaves a 0.60-meter high table
with an initial horizontal velocity of
2.4 m/s. Predict the time required for the
pool ball to fall to the ground and the
horizontal distance between the table's edge
and the ball's landing location.
Solution for Question 6
Vertical Time
∆y = vit + ½ gt2
-.60m = ½ (-9.81 m/s2)t2
-1.20m = (-9.81 m/s2) t2
t = .35 s
Horizontal Distance
x = v (t) = 2.4 m/s ( .3497s) = .84m
Question 7
A soccer ball is kicked horizontally off a
22.0-meter high hill and lands a distance of
35.0 meters from the edge of the hill.
Determine the initial horizontal velocity of
the soccer ball.
Solution Question 7
Vertical Time
∆y = vit + ½ gt2
-22m = ½ (-9.81 m/s2)t2
-44m = (-9.81 m/s2) t2
t = 2.1 s
Initial Horizontal Velocity
vx = x/t = 35 m/2.1178s = 17 m/s
Question 8
You accidentally throw your car keys
horizontally at 8.0 m/s from a cliff 64 m
high. How far from the base of the cliff
should you look for the keys?
Question 9
A toy car runs of the edge of a table that is
1.225 m high. If the car lands 0.400 m from
the base of the table,
a. how long did it take the car to fall?
b. how fast was the car going on the table?
Question 10
Divers in Acapulco dive from a cliff that is
61 m high. If the rocks below the cliff
extend outward for 23 m, what is the
minimum horizontal velocity a diver must
have to clear the rocks?
Question 11
A dart player throws a dart horizontally at a
speed of 12.4 m/s. The dart hits the board
0.32 m below the height from which it was
thrown. How far away is the player from the
board?
Question 12
A ball thrown horizontally from a 13 m high
building strikes the ground 5.0 m from the
building. With what velocity was the ball
thrown?
Question 13
A person leaps horizontally from the top of
a tower and lands 17.0 m from the base of
the tower. If the speed at which the person
was projected was 9.50 m/s, how high is the
tower?
Monkey & Hunter Experiment
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vectors/pap.html
PROJECTILE MOTION
Projectile Motion
Projectile Motion
Velocity
Projectile Motion
Projectile Motion
Projectile Motion
Trajectory
Up to a 45 degree angle – greater
distance at greater angle
After 45 degree – greater height, less
distance
45 degree gives the greatest distance
Any two angles that add to 90 degrees
will hit the same place
Projectile Motion
Air Resistance
Monkey and the Hunter
http://physics.bu.edu/~duffy/semester1/c04_
monkeyhunter.html
http://cccmkc.edu.hk/~kei-yhk/vx_vy1.html
Analysis of Projectile Motion
Motion
Horizontal
Vertical
Initial
Velocity
v cos
v sin
Acceleration
0
-g
(-ve for
downwards)
Time taken
t
t
Distance
moved
Final
Velocity
x = vt cos y = vt sin gt2/2
v cos
v sin - gt
Graphs for a Projectile Motion
Range R
v sin 2
R
g
2
Simulation
Terminal Velocity
Sky-diver in free fall
Terminal Velocity
Abc
Terminal velocity is reached when f(v) = g
Terminal Velocity: v-t Graph