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Physics 12
Projectile Motion
Clip of the day:
minuteEarth instead today! http://www.youtube.com/watch?v=PRbVISZ3Gc4 http://www.youtube.com/watch?v=uWZwncX_4hk
Projectile Motion
We know that an object (in the absence of air resistance) that is launched at a given angle should follow a parabolic path
We need to break the velocities into x and y components
Review: Projectile Motion – Horizontal Launch
An object that is launched horizontally will have no initial velocity in the y direction so the entire initial velocity will be in the x direction
At this point, we are able to treat the projectile using our two equations of motion
0
00
2
2vtatv
dtvta
td
Recall in 1D:
EQUATIONS OF MOTION
However, in 2D, we need four EOM’s to separate the horizontal and vertical components of motion:
yyy
yyy
y
xxx
xxx
x
vtatv
dtvta
td
vtatv
dtvta
td
0
00
2
0
00
2
2
2
Projectile Motion – Simplified EOMs
Considering there is no acceleration in x and the acceleration in y is g (-9.81m/s2)
yyfy
yfy
xxfx
xfx
dtvtg
d
vtgv
dtvd
vv
00
2
0
00
0
2
Review: Example 1
A cannonball is fired horizontally from the top of a 50.m high cliff with an initial speed of 30.m/s. Ignoring air resistance, determine the following:a. How long it takes to strike the groundb. How far from the base of the cliff it strikes the groundc. How fast it is travelling when it strikes the ground
Start with y position equation (4)
Sub in known information (h=50.m) and solve for time
st
st
sm
mt
ttsm
m
dtvtg
td yyy
2.3
1.10
/81.9
).50(2
002
/81.9.50
2)(
22
22
22
00
2
Now use x position equation (2) Sub in time and known
information (t=3.2s, vox=30.m/s) and solve for dx
msd
ssmsd
dtvtd
x
x
xxx
96)2.3(
0)2.3(/.30)2.3(
)( 00
Finally we will use equations (1) and (3)
Sub in time and solve for velocity
smsv
ssmsv
vtgtv
smv
vtv
y
y
yy
x
xx
/31)2.3(
0)2.3(/81.9)2.3(
)(
/.30
)(
2
0
0
Now, we employ trigonometry and Pythagorean Theorem to solve for the final velocity
o
o
y
x
smv
sm
sm
smv
smsmv
smsv
smv
46,/43
46
/.30
/31tan
/43
)/31()/.30(
/31)2.3(
/.30
1
22
vx
vy
vr
Non-horizontal launch:
Watch video and notice what happens to the velocities http://www.youtube.com/watch?v=hlW6hZkgmkA
Notice: Vx does not change and Vy goes from positive to zero at vertex to negative
Example 2:
A golfer uses a club that launches a golf ball at a 15° angle at a speed of 45m/s. Determine the following:a. The time the golf ball is in the airb. The horizontal distance the ball travelsc. The velocity as it strikes the groundd. The maximum height the ball attains
Start with the position in the y equation
Include the initial velocity in the y term
Solve for when the position in the y is equal to zero
st
smtsm
tsmtsm
tsmtsm
dtvtg
td oyoyy
4.2
)15sin(/45/91.4
)15sin(/45/91.4
0)15sin(/45/91.40
2)(
2
22
22
2
Use the time from the previous question and the position in the x equation
Solve for the range (horizontal distance)
mxsd
ssmsd
dtvtd
x
x
oxoxx
2100.1)4.2(
0)4.2)(15cos(/45)4.2(
)(
Solve for the velocity in the x and y direction
smsv
smssmsv
vtgtv
smv
smv
vtv
y
y
oyy
x
x
oxx
/12)4.2(
)15sin(/45)4.2(/81.9)4.2(
)(
/44
)15cos(/45
)(
2
Use Pythagorean Theorem and Trig to solve for final velocity
o
o
y
x
smv
sm
sm
smv
smsmv
smsv
smv
15,/45
15
/44
/12tan
/45
)/12()/44(
/12)4.2(
/44
1
22
Max height will occur when y velocity is equal to zero. Solve for time and then sub into y position equation
msd
ssmssmsd
dtvtg
td
st
smtsm
vtgtv
y
y
oyoyy
oyy
9.6)2.1(
)2.1)(15sin(/45)2.1(/91.4)2.1(
2)(
2.1
)15sin(/45/81.90
)(
22
2
2
1. A boy kicked a can horizontally from a 6.5 m high rock with a speed of 4.0 m/s. How far from the base of the rock the can land?
2. 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.
3. A cannon that launches a cannon ball at a 26° angle at a speed of 33m/s. Determine the following:a. The time the cannon ball is in the airb. The horizontal distance the cannon ball travelsc. The velocity as it strikes the groundd. The maximum height the cannon ball attains
Page 536o Questions 1-8 ANSWERS: 1) 5.9 m 2) 16.5 m/s
3) a) 2.85sec b)86m c)33m/s,334° d) 10m
A FEW MORE TO TRY:
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