Transcript 2011 projectile motion

Once
a difficult problem for
cannoneers
If
the vertical and horizontal
components are separated, the
problem becomes simple.
Horizontal
– simple as a
ball rolling across the table
Use equations like x = v x t
Vertical
– simple as a free
falling gravity problem
Use equations ∆y = vi t +
(1/2)gt2
and vf = vi + gt
NOTE:
Air and other resistance is
being ignored for the time
being.
 Consider
these diagrams to answer the
following questions:
 a. Which shows the initial velocity?

a
 b. Which shows the acceleration vector?

b
What determines the time for the cannon ball to hit the ground?
 The
time for the cannon ball to hit the
ground is determined by the height from
which it drops.
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

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
 http://www.physicsclassroom.com/mme
dia/vectors/pap.html
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)
How high is the 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
At what speed will it hit the ground?
 Horizontally
 30
speed is constant.
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
 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?



 ∆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?


 ∆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
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.
 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
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.
 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
 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?
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
table?
fast was the car going on the
 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?
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?
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?
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?
Problem

Solving Approach
Carefully read the problem. List known and unknown
information
 For convenience sake, make a table with horizontal
information on one side and vertical information on the other
side.
 Identify the unknown quantity which the problem
requests you to solve for.
 Select either a horizontal or vertical equation to
solve for the time of flight of the projectile.
 With the time determined, use one of the other equations to
solve for the unknown. (Usually, if a horizontal equation is used to
solve for time, then a vertical equation can be used to solve for the
final unknown quantity.)
Velocity
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
Sky-diver in free fall
 Abc
Terminal velocity is reached when f(v) = g