Thurs Sept 10
Today:
• Projectile motion
• Soccer problem
• Firefighter example
• Assign 2 – Friday
• SI Sessions: Morton 227
– Mon 8:10-9:10 PM
– Tues 8:10-9:10 PM
– Thur 7:05-8:05 PM
• Read – Read
• Draw/Image – lay out coordinate system
• What know? Don't know? Want to know?
• Physical Processes? Laws? Conditions?
• Valid Relationships/Formulas
vx = v0 x
• Solve
• Break problems into
components
x = x0 + v0 xt
y = y0 + vyt
1
vy = (v0 y + vy )
2
vy = v0 y + ayt
1 2
y = y0 + v0 yt + ayt
2
vy2 = v02 y + 2ay (y− y0 )
• Link with time
http://capa2.phy.ohiou.edu/res/ohiou/physlets/2dkinematics/proj2.html
Adding 2 Vectors Graphically - Lab
R
A
A
A
Tail-to-head
method
B
B
R is Resultant – the result of adding a
group of vectors
Parallelogram method
• works for 2 vectors
• Create parallelogram out of two vectors
• Vector from the two tails to the opposite
corner is the resultant
R
A
B
Lab Next Week: Forces and Equilibrium
•
•
•
•
Equilibrium: forces balanced, acceleration = 0
If a=0, forces are balanced
Weight: Force due to gravity: Fg = mass*g
Balance forces: both in x and y
Definitions: Average velocity & acceleration
Chapter 3
Constant Acceleration in 2 Dimensions
• If acceleration constant: 2 sets of equations for horizontal and vertical motion
g= 9.8 m/s2 downward
x = x0 + vxt
y = y0 + vyt
1
vx = (v0 x + vx )
2
1
vy = (v0 y + vy )
2
vx = v0 x + axt
vy = v0 y + ayt
1 2
x = x0 + v0 xt + axt
2
1 2
y = y0 + v0 yt + ayt
2
2
x
2
0x
v = v + 2ax (x − x0 )
vy2 = voy2 + 2ay (y− y0 )
Projectile Motion
2
• If free-fall, acceleration 9.8 m/s downward AND
• If choose y-axis vertical (+y up typically)
2
2
• ax = 0 m/s , ay = 9.8 m/s downward
http://capa2.phy.ohiou.edu/res/ohiou/physlets/2dkinematics/proj2.html
https://phet.colorado.edu/en/simulation/projectile-motion
x = x0 + vxt
1
vx = (v0 x + vx )
2
vx = v0 x + axt
1 2
x = x0 + v0 xt + axt
2
2
x
2
0x
v = v + 2ax (x − x0 )
y = y0 + vyt
vx = v0 x
x = x0 + v0 xt
1
vy = (v0 y + vy )
2
vy = v0 y + ayt
1 2
y = y0 + v0 yt + ayt
2
vy2 = v02 y + 2ay (y− y0 )
"Reading the Problem" Verbal and visual clues
Common theme: How do I find extra information when I don't think I
have enough? How do I know what information refers to what
variable?
- Speed refers to magnitude of velocity (v).
- Look for clues like 'horizontal/vertical', look at drawing, look at
path.
- If bullet leaving a horizontal gun barrel, what's the initial vertical
component of the velocity?
- Be careful not to assume things you can’t:
remember “just after leaves”, “just before hits”
As soon as extra force, acceleration no longer 9.8 m/s2
Water Tank
Water leaves the tank traveling
horizontally at a speed of 4 m/s.
The spigot is 1 m above the
ground.
What are the values for the following quantities?
vx0, vy0, ax, ay?
vx0 = 4 m/s
vy0 = 0 m/s
ax = 0 m/s2
ay = 9.8 m/s2 downward (which in this case is negative)
How much time to hit ground?
Ignore horizontal components
1 2
y = y0 + v0 yt + ayt
2
1
−1.0m = (0m/s)t + (−9.8m/s2 )t 2
2
Example: Soccer Ball
! Sect 3.3
http://capa2.phy.ohiou.edu/res/ohiou/physlets/2dkinematics/proj2.html
A soccer ball is kicked on a level playing field. It's initial velocity is
20.0 m/s at an angle of 30° above the horizontal.
• What is the maximum height the ball reaches?
• How much time is the ball in the air?
• What is the final speed of the ball just before it hits the ground?
• How far away does it hit the ground? (what is the range)
vx = v0 x
x = x0 + v0 xt
vy = v0 y + ayt
y = y0 +
1 2
y = y0 + v0 yt + ayt
2
1
v0 y + vy )t
(
2
vy2 = v02 y + 2ay (y− y0 )
A soccer ball is kicked on a level playing field. It's initial velocity is
20.0 m/s at an angle of 30° above the horizontal.
• What is the maximum height the ball reaches?
(1) Drawing
(3) Initial/Final
+y
Initial: just after ball leaves foot
+x
Final: ball at max height
x=0,y=0
(2) Axes and origin
(4) Known/Unknown – Split into Horizontal and Vertical
y=?
x=?
vy0 = 20 m/s sin(30º)
vx0 = 20 m/s cos(30º) t = ?
= +10m/s
= 17.3 m/s
vy = 0 m/s
2
ax = 0 m/s
ay = 9.8 m/s2 down
= -9.8 m/s2
(5) Const accel: Solve
vy2 = vy02 + 2ayy
02 = (10 m/s)2 + 2(-9.8 m/s2)y
y = 5.1m
Soccer Ball:
• How much time is the ball in the air?
• How far away does it hit the ground?
Same drawing and coordinate system as last part
(3) Initial/Final
y = vy0 t + ½ayt2
Initial: just after ball leaves foot
0 = (10m/s)t +½ (-9.8m/s2)t2
Final: just before hits ground
0 = t(10m/s – 4.9t)
(4) Known/Unknown – Split into Horizontal and Vertical
x=?
vx0 = 20m/s cos(30º)
= 17.3m/s
ax = 0 m/s2
t=?
(5) Const accel: Solve
y= 0m
vy0 = 20m/s sin(30º)
= +10m/s
vy =?
ay = 9.8 m/s2 down
= -9.8 m/s2
t = 0 or 2.04s
x = vx0 t = 17.3m/s (2.04s)
= 35.3 m
Soccer Ball:
What is the final speed of the ball just before it hits the ground?
Same drawing and coordinate system as last part
(3) Initial/Final
To find magnitude of velocity,
need vx and vy
Initial: just after ball leaves foot
Final: just before hits ground
(4) Known/Unknown – Split into Horizontal and Vertical
x=?
vx0 = 20m/s cos(30º)
= 17.3m/s
ax = 0 m/s2
t=?
(5) Const accel: Solve
vx = vx0 = 17.3 m/s
y= 0m
vy0 = 20m/s sin(30º)
vy = vy0 + ay t
= +10m/s
vy =?
vy= -10 m/s
vy = 10m/s + (-9.8m/s2)(2.04s)
ay = 9.8 m/s2 down
= -9.8 m/s2
17.3m/s
θ
10m/s
v = v x2 + v y2 = 17.32 + 10 2 = 20m/s
θ = tan-1(10/17.3) = 30º below horizontal
A firefighter is spraying water on a building. Water leaves the hose at 35 m/s and an
angle of 30° above the horizontal. The nozzle of the hose is 1.0 m above the ground,
and the building is 22 m away. The building is 40 m tall. What are the horizontal (x) and vertical (y) components of the initial velocity (just
after the water leaves the hose)?
(1) vx0 = 30.3 m/s; vy0= 30.3 m/s
(2) vx0 = 17.5 m/s; vy0= 17.5 m/s
(3) vx0 = 17.5 m/s; vy0= 30.3 m/s
(4) vx0 = 30.3 m/s; vy0= 17.5 m/s
(5) vx0 = 35.0 m/s; vy0= 17.5 m/s
(6) vx0 = 35.0 m/s; vy0= 35.0 m/s
35 m/s
35.0 sin(30º) =
17.5 m/s
30º
35.0 cos(30º) = 30.3 m/s
A firefighter is spraying water on a building. Water leaves the hose at
35 m/s and an angle of 30° above the horizontal. The nozzle of the
hose is 1.0 m above the ground, and the building is 22 m away. The
building is 40 m tall.
How much time does it take for the water to reach the building?
(1) 0.63s
(2) 0.73s
(3) 1.00s
(4) 1.32s
(5) 1.78s
You have enough information to determine this from the horizontal part
of the problem:
You know vx=35*cos30 and that the water travels 22 m
A firefighter is spraying water on a building. Water leaves the hose at
35m/s and an angle of 30° above the horizontal. The nozzle of the
hose is 1.0m above the ground, and the building is 22m away. The
building is 40m tall. In what vertical direction is the water traveling when it hits the
building?
(1) up
(2) level
(3) down
Find vy at t=0.72 s. If it is positive, water traveling up. If negative,
water traveling down.
vy = vy0 + at = 35*sin(30) + (-9.8)*0.72s = + 10.4 m/s
Upward
A firefighter is spraying water on a building. Water leaves the hose at
35m/s and an angle of 30° above the horizontal. The nozzle of the
hose is 1.0m above the ground, and the building is 22m away. The
building is 40m tall. What is the speed of the water just before it hits the building?
(1) 17.5 m/s
(5) 47.8 m/s
(2) 30.3 m/s
(3) 32.0 m/s
(4) 35.0 m/s
Need x and y components of speed. vx = 35*cos(30)=30.31 m/s
vy just found as 10.4 m/s
Use Pythagorean's theorem to find speed: 32.0 m/s
Velocity and Speed
35 m/s
17.5 m/s
30º
30.3 m/s
•
•
•
•
•
Speed is magnitude (size) of velocity vector. Here it is 35 m/s
Velocity would be 35 m/s 30º above the horizontal
Vertical component of velocity would be +17.5 m/s
Horizontal component of velocity would be +30.3 m/s
If you need the speed, you probably need to find the two
components and reconstruct it
• At the beginning of the problem, you'll likely have to break
the velocity into components.
Three projectiles are launched as described. Which one reaches the
highest altitude?
1.
2.
3.
4.
A 5 kg ball launched straight up at 20 m/s
A 4 kg ball launched at 24 m/s at an angle 70° above the horizontal
A 2 kg ball launched at 27 m/s at an angle 50° above the horizontal
All three reach the same maximum altitude
Compare vertical components of velocity. Largest vy will go the
highest.
A water balloon is launched at a speed of 11 m/s from ground level at
an angle of 50 degrees above the horizontal.
The water balloon is launched toward an innocent bystander who is
1.5 m tall and standing 2 m away.
a) By how much does the water balloon clear the person's head?
b) How tall can a person be and still have the balloon clear?
c) What is the tallest a person can be and still have the balloon clear
if they can stand where ever they want?
d) How fast is the balloon traveling when it is directly over the
person's head in part (a)?
(a) 0.49 m
(b) 1.99 m
(c) 3.63 m
(d) 9.06m/s
Student A is hanging from a tree. Student B is on a local hillside
aiming a water balloon directly horizontally at Student A. Student B
lets out a loud scream and releases the water balloon. Student A lets
go of the branch at the same time the water balloon is released. The
water balloon will:
1. pass above Student A
2. hit Student A
3. pass below Student A
Both student and water balloon have an initial velocity of zero and fall at
same rate.
Hot Wheels on Coffee Table
Ryan pushes his Hot Wheels car horizontally off the coffee table with
just the right speed to stick it in between the cushions on our living
room chair.
The chair is 0.35 m from the edge of the table. The crack between the
cushions is 0.05 m below the level of the coffee table.
1. How much time is the car in free fall?
2. How fast must he push the car to hit the middle of the crack?