scrambled eggs see eye to eye up to no good pipe down or downpipe
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scrambled eggs
see eye to eyeup to no good
pipe down or downpipe
3.2 Equations of Kinematics in Two Dimensions
• Equations of Kinematics
tvvx o 21
221 attvx o
atvv o
axvv o 222
3.2 Equations of Kinematics in Two Dimensions – the horizontal component
tavv xoxx tvvx xox 21
xavv xoxx 222 221 tatvx xox
3.2 Equations of Kinematics in Two Dimensions – the vertical component
tavv yoyy
221 tatvy yoy
tvvy yoy 21
yavv yoyy 222
3.2 Equations of Kinematics in Two Dimensions• The x part of the motion occurs exactly as it
would if the y part did not occur at all, and vice versa.
Projectile MotionChapter 3Section 3
What is a projectile?
• When a long jumper approaches his jump, he runs along a straight line which can be called the x-axis.
• When he jumps, as shown in the diagram below, his velocity has both horizontal and vertical components
What is a projectile?• Projectiles are objects in free fall that have an
initial horizontal velocity.• Free fall implies that the only force acting on the object
is gravity.• Initial horizontal velocity results in motion in two
dimensions.
• Examples of projectiles include• An arrow flying towards a target• A baseball thrown in the air at an angle other than 90o,
unless they are thrown straight up.
Projectile motion
• Projectiles follow parabolic trajectories.• Trajectory is the path of a flying object.
• The horizontal and vertical motions are independent.• We can write separate equations of motion for each
direction. • Horizontal motion on the x-axis and • the vertical motion on the y-axis
What is a projectile?• If a ball is thrown horizontally, (yellow) and
another ball is dropped (red)-----they will both• HIT THE GROUND AT THE SAME TIME!!!!!!!!• We shall do a demo of this
Solving Projectile Motion Problems
• You need to separate the motion down into 2 parts!
• The first part is the horizontal movement• The second part is the vertical movement• Keep the two sets of information separate
using a table when solving such problems
Horizontal motion equations and constants
ΔtvΔx ox
projectile a ofcomponent vertical theaffectsonly gravity 0
)( 221
x
xox
a
tatvx
xoxx
xoxx
vvv
tavv
constant xv
Vertical motion equations and constants
Δt
vvg oyy
tgvv oyy
221
sm
221
))((
81.9
)(
2
tgtvy
ga
tatvy
oy
y
yoy
Calculating time of fall• If a projectile is launched horizontally, this tells
us that the object has an initial horizontal velocity (ie: in the x-axis)
• For such projectiles, the initial vertical velocity is zero (ie. in the y-axis) (voy=0)
g
yt
tgy
tgtvy oy
2
))((
))((2
21
221
Example 1
• People in movies often jump from buildings into pools. If a person jumps from the 10th floor (30.0 m) to a pool that is 5 m away from the building, with what initial horizontal velocity must the person jump?
Step 1: Write down Givens and unknowns.
• You will have to separate your y-components from your x-components using a table
X- Component Y - Component
Dx = 5.0 m (from building to pool) Dy = -30.0 m (negative due to downward direction of fall)
vox = ? voy = 0 (horizontal projectile)
ax = 0 ay = -9.81 m/s2
Step 2: Write down equations that might solve it.
g
yt
2
• We can use the y equation to solve for the time of the jump.
• We then use this calculated time in the x equation to solve for the initial velocity.
• These are the two common equations you will use in this section!!!
tvx ox
Step 3: Carry out the solution.Time for the jump
• Using the y equation to solve for the time of the jump.
g
yt
2
s 47.2 9.81-
) 2(-30
t
t
Step 4: Carry out the solution.Using our time to calculate the initial horizontal velocity
• Using the x equation to solve for the initial horizontal velocity.
m/s 0.2s 47.2
m 0.5
ox
ox
vt
x
tvx
Practice Problem 1
• An autographed baseball rolls off of a 0.70m high desk. And strikes the floor 0.25m away from the desk. How fast did it roll off the table?
Practice Problem # 1Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table)
X- Component Y - Component
Dx = 0.25 m (from base of desk) Dy = -0.7 m (negative due to downward direction of fall)
vox = ? voy = 0 (horizontal projectile)
ax = 0 ay = -9.81 m/s2
Step 2: Write down equations that might solve it.
g
yt
2
• We can use the y equation to solve for the time of the jump.
• We then use the time in the x equation to solve for the initial velocity.
tvx ox
Step 3: Carry out the solution. Time for the jump
• Using the y equation to solve for the time of the jump.
g
yt
2
s 378.0 9.81-
) 2(-0.7
t
t
Step 3: Carry out the solution.Using our time to calculate the initial horizontal velocity
• Using the x equation to solve for the initial velocity.
m/s 0.66
s 378.0
m 25.0
ox
ox
vt
x
tvx
Let us review the concepts for Horizontally launched projectiles
• Acceleration due to gravity does not affect horizontal motion
• Initial horizontal velocity = final horizontal velocity•
• Initial vertical velocity = 0m/s•
3.3 Projectile Motion
The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. a) Determine the time required for the Care package to hit the ground.b) What are the magnitude and direction of the final velocity of the Care
package?
Practice Problem # 2 - A Falling Care Package
Practice Problem # 1Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table)
X- Component Y - Component
Dx = Dy = -1050.0 m (negative due to downward direction of fall)
vox = 115 m/s voy = 0 (horizontal projectile)
ax = 0 ay = -9.81 m/s2
Step 2: Write down equations that might solve it.
• We can use the y equation to solve for the time of the jump.• Since
g
yt
tgy
tgtvy oy
2
))((
))((2
21
221
Step 3: Carry out the solution.Time for the jump
• Therefore, using the modified y equation to solve for the time of the jump.
g
yt
2
s 6.14 9.81-
) 2(-1050
t
t
3.3 Projectile Motion
b) What are the magnitude and direction of the final velocity of the Care package?
Practice Problem # 2 - A Falling Care Package
Practice Problem # 1Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table)
X- Component Y - Component
Dx = Dy = -1050.0 m (negative due to downward direction of fall)
vox = 115 m/s voy = 0 (horizontal projectile)
ay = -9.81 m/s2
t = 14.6 s t = 14.6 s
Step 2: Write down equations that might solve it.
• We then use the time in the y equation to solve for the final vertical velocity.• Since
tavv yoyy
3.3 Projectile Motion
s 6.14sm80.90 2
tavv yoyy
sm143
Step 3: Carry out the solution.Final vertical velocity of the package
Step 4: Carry out the solution.Magnitude and direction of final velocity
22yx vvv
)143(11522
v
smv /184
115
143tan 1
2.51
CLASS ASSIGNMENT NOT H/W
I will not accept late work
3.3 Projectile Motion
Conceptual Example 5 I Shot a Bullet into the Air...
Suppose you are driving a convertible with the top down.The car is moving to the right at constant velocity. You pointa rifle straight up into the air and fire it. In the absence of airresistance, where would the bullet land – behind you, aheadof you, or in the barrel of the rifle?Explain
Section 3. Cont.
Projectiles at an Angle
Projectiles Launched at an angle
As shown below, projectiles launched at an angle have an initial vertical and horizontal component of velocity.
• Suppose the initial velocity vector makes an angle θ with the horizontal • The sine and cosine functions can be used to find the horizontal
and vertical components of the initial velocity
Projectiles Launched at an angle
3.3 Projectile Motion
Example 6 The Kickoff
• A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following:
a) determine the maximum height that the ball attains.
Step 1: Write down what you know and don’t know (and resolve the vector provided)
X – Component Y - Component
ax = 0m/s2 ay = -9.81 m/s2
Dy = ?
Dt = ?Dt = ?
)cos(oox vv )sin(ooy vv
Dx = ?
=22 m/s
=40o
Step 2: Decide on a plan or equation to use.
• For the horizontal velocity:
• For the vertical velocity:
sm1.1440sinsm22sin ooy vv
sm9.1640cossm22cos oox vv
Step 3: Carry out the plan.
y ay vy voy t? -9.81 m/s2 0 14.1 m/s
yavv yoyy 222 y
oyy
a
vvy
2
22
m 1.10
sm81.92
sm1.1402
2
y
3.3 Projectile Motion
Example 6 The Kickoff
• A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following:
a) The maximum height that the ball attains = 10 mb) The time between kickoff and landing
Step 1: Write down what you know and don’t know (and resolve the vector provided)
X – Component Y - Component
ax = 0m/s2 ay = -9.81 m/s2
Dx = ? Dy = 10.1 m
Dt = ? Dt = ?
smvox /9.16 smvoy /1.14
40ovoy
vox
V i = 22 m
/s
Step 2: Decide on a plan or equation to use.
• We can calculate tine for the vertical displacement:
tavv yoyy
• For the vertical displacement (to maximum height):
• = 2.87 s
tavv yoyy
a
vvt oyy
Step 3: Carry out the plan.
2)81.9
1.140( xt
81.9
1.141.14
t
3.3 Projectile Motion
Example 6 The Kickoff
• A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following:
a) The maximum height that the ball attains = 10 mb) The time between kickoff and landing = 2.86 sc) The range ‘R’ of the projectile?
Step 1: Write down what you know and don’t know (and resolve the vector provided)
X – Component Y - Component
ax = 0m/s2 ay = -9.81 m/s2
Dx = ? Dy = 10 m
Dt = 2.86 s Dt = 1.43 s
smvox /9.16 smvoy /1.14
40ovoy
vox
V i = 22 m
/s
Step 2: Decide on a plan or equation to use.
• For the horizontal displacement:
tvtatvx oxxox 221
• For the horizontal displacement:
• = 48.5 m
Step 3: Carry out the plan.
ssm 87.2/9.16
tvtatvx oxxox 221
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