projectile motion chapter 3.3. objectives recognize examples of projectile motion describe the path...
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Projectile Motion
Chapter 3.3
Objectives
Recognize examples of projectile motion
Describe the path of a projectile as a parabola
Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion
Projectile Motion
How can you know the displacement, velocity and acceleration of a ball at any point in time during its flight?
Use the kinematic equations of course!
Vector Components p.98 (Running vs Jumping)
While jumping, the person is moving in two dimensionsTherefore, the velocity has two components. Vy
Vx
While running, the person is only moving in one dimensionTherefore, the velocity only has one component.V
Definition of Projectile Motion
Objects that are thrown or launched into the air and are subject to gravity are called projectiles
Examples?– Thrown Football, Thrown Basketball, Long
Jumper, etc
Path of a projectile
Neglecting air resistance, the path of a projectile is a parabola
Projectile motion is free fall with an initial horizontal velocity
At the top of the parabola, the velocity is not 0!!!!!!
Vertical and Horizontal Motion
Horizontal Motion Vertical Motion
Velocity = Vx
Displacement = Δx
Velocity = Vy
Displacement = Δy
Because gravity does not act in the horizontal direction, Vx
is always constant!
Gravity acts vertically, therefore a = -9.81 m/s2
Equations for projectiles launched horizontally
Horizontal Motion Vertical Motion
Δx=Vxt
Vx is constant!
a=0
Vy,i=0 (initial velocity in y direction is 0)
ayayayvv iyfy 22022,
2,
atatatvv iyfy 0,,
222, 2
1
2
10
2
1atatattvy iy
Revised Kinematic Eqns for projectiles launched horizontally
Horizontal Motion
Vertical Motion
tvx xyav fy 22
,
atv fy ,
2
2
1aty
Finding the total velocity
Use the pythagorean theorem to find the resultant velocity using the components (Vx and Vy)
Use SOH CAH TOA to find the direction
Vx
Vy
V
Example p. 102 #2
A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. What was the cat’s speed when it slid off the table? What is the cat’s velocity just before it hits the ground?
What do we know and what are we looking for?
2.2m
Δx= 2.2 m Δy= -1.0m (bc the cat falls down)
1.0
mVx= ?????
What are we looking for??
How do we find Vx?
Equation for horizontal motion:
We have x…so we need t.
How do we find how long it takes for the cat to hit the ground?
Use the vertical motion kinematic equations.
tvx x
Vertical Motion
Δy= -1.0m a=-9.81 m/s^2 What equation should we use?
Rearrange the equation, to solve for t then plug in values.
2
2
1aty
s
smm
a
yt 45.
81.9
)0.1(22
2
Horizontal equation
Rearrange and solve for Vx:
Cat’s Speed is 4.89 m/s
tvx x
s
m
s
m
t
xvx 89.4
45.
2.2
Cliff example
A boulder rolls off of a cliff and lands 6.39 seconds later 68 m from the base of the cliff. – What is the height of the cliff?– What is the initial velocity of the boulder?– What is the velocity of the boulder just as it strikes
the ground?
How high is the cliff?
Δy= ? a=-9.81 m/s2
t = 6.39 s Vx=?
Vy,i = 0 Δx= 68 m
2
2
1aty
ms
maty 200)39.6)(81.9(
2
1
2
1 22
2
The cliff is 200 m high
What is the initial velocity of the boulder?
The boulder rolls off the cliff horizontally
Therefore, we are looking for Vx
tvx x
s
m
s
m
t
xvx 6.10
39.6
68
Important Concepts for Projectiles Launched Horizontally
Horizontal Components Vertical Components
Horizontal Velocity is constant throughout the flight
Horizontal acceleration is 0
Initial vertical velocity is 0 but increases throughout the flight
Vertical acceleration is constant: -9.81 m/s2
Projectiles Launched at An Angle
Projectiles Launched Horizontally
Projectiles Launched at an Angle
•Vx is constant
•Initial Vy is 0
•Vx is constant
•Initial Vy is not 0
Vi
Vx,i
Vy,i
θ
Vi = Vx
Components of Initial Velocity for Projectiles Launched at an angle
Use soh cah toa to find the Vx,i and Vy,i
i
ix
v
v
h
a ,)cos(
i
iy
v
v
h
o ,sin
Vi
Vx,i
Vy,i
θ
cos, iix vv
sin, iiy vv
Revise the kinematic equations again
Horizontal Motion Vertical Motion
2
2
1sin tatvy i
cos, iixx vvv
tvtvx ix cosyavyavv iiyfy 2))sin((2 22
,2
,
atvatvv iiyfy sin,,
Example p. 104 #3
A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high was the tallest spot in the ball’s path?
What do we know?
Δx= 42.0 m
θ= 25°
Vi= 23.0 m/s
Vy at top = 0
Δt=?
Δy=?42.0 m
25°
What can we use to solve the problem?
Find t using the horizontal eqn:
Δx=vxΔt = vicos(θ)t
How to find Δy?– Vy,f = 0 at top of the ball’s path
– What equation should we use?
s
sm
m
v
x
v
xt
ix
0.225)(cos23(
0.42
cos
yavv iyfy 22,
2,
m
sm
sm
a
v
a
vvy iiyfy 8.4
81.92
25sin)23(0
2
)sin(0
2
2
2
22,
2,
Cliff example
A girl throws a tennis ball at an angle of 60°North of East from a height of 2.0 m. The ball’s range is 90 m and it is in flight for 6 seconds.
– What is the initial horizontal velocity of the ball?– What is the initial vertical velocity of the ball?– What is the total initial velocity of the ball?– How high above the initial position does the ball get?– What is the vertical velocity of the ball 2 seconds after it is
thrown?
What is the initial horizontal velocity of the ball?
Δx= 90 m Θ=60° Total time= 6 s Horizontal velocity is
constant: Vx
tvx x
East 156
90
s
m
t
xVx
What is the initial vertical velocity of the ball?
iVx
iVy
,
,tan
North 2698.25)15)(60tan(, s
m
s
mV iy
Vx,i
Vy,i
θ
Vi
What is the total initial velocity of the ball?
2,
2,
2iyixi VVV
Vx,i
Vy,i
θ
Vi
East ofNorth 60at 302615 22 s
mVi
How high above the ground does the ball get?
At the top of the parabola, Vy is 0…so use the revised kinematic equations
Add 2m to get the height above the ground: 36.65 m
yavv iyfy 22,
2,
m
s
ma
vvy iyfy 45.34
81.92
260
22
22,
2,
What is the vertical velocity of the ball 2 seconds after it is thrown?
Vy,i=+26 m/s a= -9.81 m/s2
t = 2 seconds
s
ms
s
m
s
matvv iyfy 4.6)2)(81.9(26
2,,
Important Concepts for Projectiles Launched at an Angle
At the top of the parabola, neither the object’s velocity nor it’s acceleration is 0!!!!!– Only Vy is 0
– Vx is constant throughout the flight
– Horizontal acceleration is always 0– Vertical acceleration is always -9.81 m/s2