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Chapter 7
Forces in Two Dimensions
equilibrium, projectile and circular motion
Two problems*
Table lift demo: “Force
applied at an angle to the
direction of motion yields
less of an effect on the
object’s motion than force
applied in a straight line
with the motion.”
Draw a vector diagram to
show the forces and explain
the experience of the test
subject.
Derek left his physics
book on top of a drafting
table. The table is inclined
at a 35° angle. Find the
net external force acting
on the book, and
determine whether the
book will remain at rest in
this position.
(given: Fg = 22N. Ff = 11N, FN
= 18N)
*answers on Thursday
the Energy and force story
a quick review

what you already know
review problem
An 1850 kg car is moving right at a constant
velocity of 1.44 m/s. What is the net force on the
car?
0
as long as speed is constant
review problem .
What net external force is required to give a
25 kg suitcase an acceleration of 2.2 m/s2 to
the right?
a = F/m
2.2 = F/25
55 N
review problem
A boat moves through the water with two
forces acting on it. One is a 2.10 X 103 N
forward push by the motor, and the other is a
1.80 X 103 N resistive force due to the water.
What is the acceleration of the 1200 kg boat?
Fnet = ma
2.10 X 103 - 1.80 X 103 = (1200)(a)
.3 X 103 /1200 = a
0.25 m/s2
review problem
A dog pulls on a pillow with a force of 5 N at an
angle of 37° above the horizon. Find the x and y
components of this force.
5N
37°
x = 5cos37
3.9 N
y = 5sin37
3N
Sec. 7.1 forces in 2 dimensions
An object is in equilibrium
when the net force on it is 0.
an object is motionless
When in equilibrium
an object moves with
constant velocity
Equilibrium – Occurs when the resultant of three or
more forces equals a net force of zero.
If A+B+C=0
B
Then A, B & C
are at
equilibrium
A
C
B
C
A

If an object is in equilibrium,
all of the forces acting on it
are balanced and the net force
is zero.
The 2 forces acting in the
same direction are additive.

If the forces act in two
dimensions, then all of the
forces in the x-direction and ydirection balance separately.


It is much more difficult for
the gymnast to hold his arms
out at a 45-degree angle.
each arm must still support
350 N vertically to balance
Thisforce
is theofsame
as in our table
the
gravity.
demonstration from yesterday.

Use
the
vertical
forceabout
vector
Using
what
you know
(y-component)
to find the
forces and equilibrium,
draw a
total
force
in each
of the
vector
diagram
to show
the
gymnast’s
(xthe
forces and arms
explain
experience of the test subject.
component).
350N each arm
vertically
The ball carrier and
tackler collide and are
still moving forward.
How can a 2nd tackler
stop the forward
motion (produce
equilibrium)?
When 2 forces are
acting on an object
and the sum is not 0
– a 3rd force can be
added that produces
equilibrium.
*a force that produces equilibrium
finding the equilibrant:

To find this force, first find the sum of the two forces
already being exerted on the object.

This single force that produces the same effect as
the two individual forces added together, is the
resultant force.
15
finding the equilibrant of A and B

The equilibrant force is one with the same magnitude
as the resultant force, but in the opposite direction.
Forces are balanced
and the net force is 0
16
equilibrant in the
opposite direction
Review: Components of vectors
Going backwards to resolve a resultant vector into
the vectors that produced it (vector resolution)
+y
Ry
q

Rx
17
+x
To find the x component, Rx:
cos θ = a/h
cos 40 = Rx/5.0km  Rx = 5.0km (cos 40)
Rx = 3.8 km
To find the y component, Ry:
sin θ = o/h
sin 40 = Ry/5.0km  Ry = 5.0km (sin 40)
Ry = 3.2 km
+y
3.2 km R y
q = 40o
Rx
3.8 km
18
+x
Example problem pg. 151
A 168N sign is hanging motionless by two ropes
that make 22.5° angles with the horizontal. What
is the tension in the ropes?
22.5
22.5
First identify all the forces
involved.
Equilibrium Problem Solving




The sum of the forces must equal zero
Draw all forces on the object
Break forces into (x,y) components
Combine forces that are in the same direction
Fa
Fb
Break forces into horizontal
and vertical components.
Fw
Fa
Fb
Fay
Fby
Fbx
Fax
Fw
Fa
Fb
Fay
Fby
Fbx
Fax
Fw
Fa
Fay=Fasin22.5
Fay
Fax=Facos22.5
Fax
Fb
Fby=Fbsin22.5
Fby
Fbx
Fbx=Fbcos22.5
Resolve Forces By Direction
Fa
Fb
Fay
Fby
Fbx
Fax
Fnety=0
Fnetx=0
Fw
Fnety= 0 = Fay + Fby - Fw
Fnetx= -Fax+Fbx
0 = 2Fasin22.5 - Fw
0=-Facos22.5+Fbcos22.5
2Fasin22.5 = Fw
Facos22.5=Fbcos22.5
Fa = 168N/2sin22.5
Fa=Fb
Fa = 220N = Fb
Practice problems 1-4, pg. 151/152
problem 2, pg 151:
An 8.0 N weight has one horizontal rope exerting a
force of 6.0 N on it.
a. What are the magnitude and direction of the
resultant force on the weight?
b. What force (magnitude and direction) is
needed to put the weight into equilibrium?
Draw a vector diagram
(Draw all forces on the object)
Break forces into (x,y) components
Combine forces that are in the same direction
Determine the equilibrant
An 8.0 N weight has one horizontal rope exerting a force of 6.0 N on it.
a. What are the magnitude and direction of the resultant force on the weight?
FR = 10 N at 310⁰
ϴ?
6.0 N
FX2 + FY 2 = FR2
ϴy
FR ?
FR = √(6.0 N)2 + (8.0 N)2
= 10 N
8.0 N
b. What force
(magnitude and
direction) is needed
to put the weight into
equilibrium?
FE = 10 N at (307⁰ - 180⁰)
= 10 N at 127⁰
ϴR = 270⁰ + ϴy
ϴy = tan-1 (6.0/8.0) = 37⁰
ϴR = 270⁰ + 37⁰ = 307⁰
Forces on an Incline
remember:
• Define your coordinates
• Draw a FBD
• Identify all forces.
• Resolve into x- and y- components.
How do you determine if a body is in equilibrium?
• Assign a coordinate system
• Resolve all forces into their x- and y-components
• When ΣFx = 0 AND ΣFy = 0, then the object is in
equilibrium. (velocity will not change)
θ?
Fg = 22 N
Ff = 11N
FN = 18N
As shown in the diagram, there are always at least two
forces acting upon any object on an inclined plane - the
force of gravity and the normal force. The force of gravity
(weight) acts in a downward direction; yet the normal
force acts in a direction perpendicular to the surface
Remember: normal forces are always
directed perpendicular to the surface
that the object is on.
Analyzing the forces acting upon objects on inclined planes
involves resolving the weight vector (Fgrav) into two
perpendicular components - one directed parallel to the
inclined surface and the other directed perpendicular to the
inclined surface.
Together the two components replace the affect of the
force of gravity.
The perpendicular component of the force of gravity is directed
opposite the normal force (balances the normal force).
The parallel component of the force of gravity is not balanced by any
other force.
An object will accelerate down the inclined plane due to the presence
of an unbalanced force. It is the parallel component of the force of
gravity that causes this acceleration. The parallel component of the
force of gravity is the net force.
inclined plane problems can be simplified through a
useful trick known as "tilting the head."
Once the force of gravity has been resolved into its two components and the
inclined plane has been tilted, merely ignore the force of gravity (since it has been
replaced by its two components) and solve for the net force and acceleration.
The task of determining the
magnitude of the two
components of the force of
gravity is a matter of resolving
the (resultant) gravity vector into
its two components, x and y
example problem pg. 152
A trunk weighing 562 N is resting on a plane inclined
at 30°. Find the components of the weight force
parallel and perpendicular to the plane.
remember:
FN
30°
Fg
remember:
sinθ = Fg,x/Fg
Fg = W = mg
Fg,x = mg x sinθ
(562)(sin 30°)
= 281 N
cosθ = Fg,y/Fg
Fg = W = mg
Y
Fg,y = mg x cosθ
(562)(cos 30°)
= 487 N
X
the short cut equation
Once you understand how to derive the components –
use the “shortcut” equations
The equations for the parallel and
perpendicular components are:
y+
F = mg x sin θ
F = mg x cos θ
X+
F = ma
mg x sin θ = ma
mg x sin θ/m = a
g x sin θ = a
In the absence of friction and other forces the acceleration
of an object on an incline is the value of the parallel (x)
component (mg x sin θ) divided by the mass (m). This yields
the equation:
a = g x sin θ
try an example…
a 1000-kg roller coaster on
the first drop of a roller
coaster ride is at an angle
of 45°.
Determine the net force on
the riders and acceleration
of the roller coaster car.
(Assume a negligible affect of
friction and air resistance.)
Fgrav = m • g = (1000 kg) • (9.8 m/s2) =
9800 N
The parallel and perpendicular
components of the gravity force can
be determined from their respective
equations:
Fll(x) = mg • sin 45° = 6930 N
Fperpendicular = mg • cos 45° = 6930 N
The forces directed perpendicular to the
incline balance each other.
There are no other forces to
counteract the parallel
component of gravity (x).
the net force = the Fll value.
Fnet = 6930 N, down the incline (the force on the riders)
The acceleration: a = Fnet /m:
= 6930 N/1000 kg
a = 6.93 m/s2, down the incline
when other forces are involved…
In the presence of friction (or other forces), the situation is slightly
more complicated.
The perpendicular component of force
balances the normal force.
Now the frictional force must also be
considered when determining the net force of
the X (parallel) component.
The two parallel forces must be added to
determine the net force. The parallel
component and the friction force add together
to yield 5 N. The net force is 5 N, directed
along the incline towards the floor.
will the physics book slide?
FN = 18N
35°
Fg = 22 N
Known:
Fg = 22 N
Ff = 11N
FN = 18N
θ = 35°
unknown:
ΣFx = 0
ΣFy = 0
need to determine the x and y components for the Fg
Ff = 11N
FN = 18N
Fx = mg x sin θ
Fy = mg x cos θ
35°
Fy = 22 x cos 35
= 18 N
ΣFN – Fy = 0
Fg = 22 N
The parallel forces determine if the book will slide
Fx= 22 x sin 35
= 12.6 N
know Ff = 11N
ΣFx – Ff = 12.6 – 11 = Fnet of 1.6 N downward – the book will slide
try another problem
a 100-kg crate is
sliding down an
inclined plane at an
angle of 30°.
The coefficient of
friction between the
crate and the incline is
0.3.
Determine the net force and acceleration of the crate.
quick review – coefficient of friction
Ffriction = μFN
Lee Mealone is sledding with his
friends when he becomes
disgruntled by one of his friend’s
comments. He exerts a rightward
force of 9.13N on his 4.68 kg sled
to accelerate it across the snow. If
the acceleration of the sled is
0.815 m/s2, what is the coefficient
of friction between the sled and
the snow?
known: Fapplied = 9.13N,right m = 4.68 kg a = 0.815
m/s2 unknown: μ between sled and snow
FN
Ffriction = μFN
9.13N
Fg
Fg = ma
= (4.68)(9.8)
= 45.86 N
FN = Fg
Fnet = 0 because the sled has an acceleration
Fnet = (4.68)(0.815) = 3.81 N, right so Fapp > Ff
Ff = Fnet – Fapp = 3.81N – 9.13N = 5.32N, left
μ = Ff/FN = 5.32/45.86 = 0.116
one more
A 405N rightward force is used to drag a
large box across the floor with a constant
velocity of 0.678 m/s. The coefficient of
friction between the box and the floor is
0.795. Determine the mass of the box.
405 N
known:
Fapp = 405N,right μ = 0.795 a = 0
unknown: m of the box
Velocity is constant so all forces are balanced,
therefore: Fapp = Ffriction = 405 N AND FN = Fg
FN = Ff /μ = 405/0.795 = 509.43N
FN = Fg
so 509.43N = m(9.8) = 51.98 kg
back to a problem on an incline……
a 100-kg crate is sliding
down an inclined plane
at an angle of 30°.
The coefficient of friction
between the crate and
the incline is 0.3.
Determine the net force and acceleration of the crate.
First: find the force of gravity
acting on the crate and the
components of this force
parallel and perpendicular to
the incline.
100 kg
The force of gravity (W = mg) is 980 N
The components of this force are Fll(x) = 490 N (980N • sin30°)
and Fy = 849 N (980 N • cos30°).
Now the normal force can be determined to be 849 N (it must
balance the perpendicular component of the weight vector).
Determine the net force and acceleration of the crate.
Determine the force of friction
Ffrict = μFN = 0.3 x 849 N
= 255 N
The net force is the vector sum of all the forces.
The forces directed perpendicular to the incline balance;
The forces directed parallel to the incline do not balance.
Fnet = 490 N - 255 N = 235 N
acceleration is Fnet/m = 235 N/100 kg) = 2.35 m/s2.
Try another (example pg. 153)
A 62 kg person on skis is going down a hill sloped at
37°. The coefficient of kinetic friction between the
skis and the snow is 0.15. How fast is the skier
going 5.0 s after starting from rest?
Fg = mg = 62(9.8) = 607.6
Fx = sin 37° x mg = 365.7N
Fy = FN = cos 37° x mg = 485.2N
37°
Ffriction = μFN
= (.15)(485.2) = 72.78N
Fnet = Fx – Ff = 365.7 – 72.78 = 292.9N
Fnet = Fx – Ff = 365.7 – 72.78 = 292.9N
a = Fnet/m
= 292.9/62
= 4.72 m/s2
Vf = V0 + at
= (4.72)(5s)
= 23.6 m/s
Practice problems 5-8, pg. 154
Study guide problems
Worksheet problems
Projectile Motion
Sec. 7.2
The most common example of an object moving in two dimensions
is a projectile.
After a projectile has been given initial thrust, it
moves only by the force of gravity (ignoring air resistance).
If you know the initial thrust on a projectile your can figure
out its trajectory**.
** its path
types of projectiles
A projectile is any object that once projected continues in
motion by its own inertia and is influenced only by the
downward force of gravity.
Remember: A force is not required to keep an object in motion. A force
is only required to maintain an acceleration.
Gravity is the downward force on a projectile that influences
its vertical motion and causes the parabolic trajectory that is
characteristic of projectiles.
Recap:
• A projectile is an object upon which the only force is
gravity.
• Gravity acts to influence the vertical motion of the
projectile, thus causing a vertical acceleration.
• The horizontal motion of the projectile is the result of
the tendency of any object in motion to remain in motion
at constant velocity.
• Due to the absence of horizontal forces, a projectile
remains in motion with a constant velocity. Horizontal
forces are not required to keep a projectile moving
horizontally.
The only force acting upon a projectile is gravity!
There are the two components of the projectile's motion horizontal and vertical motion (they are independent of each other).
Horizontal motion: Law of
inertia (absence of gravity)
Vertical motion:
acceleration of gravity
Only a vertical force
acts on the projectile.
the projectile travels with a constant horizontal velocity and
a downward vertical acceleration.
Horizontal motion



The ball’s horizontal velocity
remains constant while it falls
because gravity does not
exert any horizontal force.
Since there is no force, the
horizontal acceleration is zero
(ax = 0).
The ball will keep moving to
the right at 5 m/sec.
Vertical motion



The vertical speed (vy) of the
ball will increase by 9.8 m/sec
after each second.
After one second has passed,
vy of the ball will be 9.8 m/sec.
After 2 seconds have passed,
vy will be 19.6 m/sec and so on.
Remember for “cliff” problems:
Horizontal and vertical motion act independently
d x = vt
vfx = vix = constant
1 2
d y = at
2
vfy = viy - gt
Projectile motion
is a combination
of horizontal and
vertical motion
but neither
affects the other
ANALYSIS OF MOTION
ASSUMPTIONS:
•
x-direction (horizontal):
uniform motion
•
y-direction (vertical):
accelerated motion
•
no air resistance
QUESTIONS:
•
What is the trajectory?
•
What is the total time of the motion?
•
What is the horizontal range?
•
What is the final velocity?
Solving projectile problems: pg 156
Horizontal motion - same as constant velocity problems (1D)
a = 0 v = d/t
dx = vt
Time*
Vertical motion – solve like a free fall problem
v = at a = v/t a = -g dy = -1/2at2
*solve for either -- same for both
How fast is the ball thrown?



Vertical dy = -1/2at2
5 = -.5(9.8)t2
t= 1 second
►
Horizontal d = vt
► v = d/t
► v= 20m/1s
► v= 20 m/s
The ball is thrown at 20 m/s
example pg. 157
A stone is thrown horizontally at 15 m/s from the
top of a cliff 44 m high.
a. How far from the cliff does it hit the ground?
b. How fast is it moving when it hits?
Know:
vx = 15 m/s
a = -g
vy = 0
a. Solve for t in the air
dy = -1/2at2
Want to know:
x when y= - 44m
v at that time
t = √-2d/a
t = √ -2(44 m) = 3 s
-9.8 m/s/s
vx
t = 3s
dx = vt
= (15 m/s)(3.0 s) = 45 m
vy
b. How fast is it moving?
vy = -gt
= (-9.8 m/s/s)(3.0s) = -29 m/s
v2net = v2x + v2y
v = √ (15 m/s)2 + (-29 m/s)2 = 33 m/s
one more example (#11, pg 158):
A steel ball rolls with constant velocity across a tabletop
0.95 m high. It rolls off and hits the ground 0.352 m from
the edge of the table. How fast was it rolling?
vy = 0
y = -1/2gt2
t = √-2y/g
t = √ -2(-.95 m) = 0.44 s
9.8 m/s/s
vx = d/t
0.95 m
v = 0.352 m/ 0.44s = 0.800 m/s
another example problem
Wile E. Coyote accidentally runs off a cliff horizontally at
8.0 m/s. The cliff is 64 m high. How far from the base of
the cliff should the road runner find him?
Formulas: vx = d/t
rearrange: d = vxt
y = -1/2gt2
t = √2y/g
solve for time: t = √2(64) / 9.8
= 3.6 s
Use time to solve for dx
d = v xt
= (8 m/s)(3.6s)
= 28.8 m
Example problem
A stunt driver steers a car off a cliff at a
speed of 20 meters per second. He
lands in the lake below two seconds
later. Find the height of the cliff and
the horizontal distance the car
travels.
1.
2.
3.

You know the initial speed and the time.
Use relationships: y = – ½ gt2 and x = vox t
The car goes off the cliff horizontally, so voy = 0. Solve:
y = – (1/2)(9.8 m/s2)(2 s)2 y = –19.6 m. (negative means the
car is below its starting point)

Use x = voxt, to find the horizontal distance: x = (20 m/s)(2 s)
x = 40 m.
A cow walks off a bridge at a speed of 2 m/s. It takes 9
seconds for the cow to land.
a)
How far away from the base of the
bridge will the cow land?
V=d/t
2 m/s = d / 9 s
d = 18 m
Recall that the cow walked at a speed of 2 m/s and
took 9 seconds to land.
b) How tall is the bridge?
d = ½ g x t2
d = ½ (10m/s2) x 9 s2
d = 405 m
While driving Eleanor his freshly boosted 1967 Shelby
Mustang, Randall “Memphis” Reines decides to
escape from Det. Castlebeck by jumping over a car
accident using the flat bed tow truck. The bed of the
tow truck is 4 m high and Eleanor is traveling at 44
m/s.

How long does it take for Eleanor to reach the ground on
the other side of the car accident?

How far horizontally from the
end of the tow truck bed did
Eleanor fly?
Gone in 60
seconds


How long does it take for Eleanor to hit the ground on the
other side of the accident?
d = ½ g x t2
t = √2 d / g
t = √ 2 (4 m)/(10 m/s2)
t = 0.9 s
How far horizontally from the end
of the tow truck bed did Eleanor fly?
v=d/t
. d = 44 m/s x 0.9 s
. d = 39.6 m
.d=vxt
Be sure to have all these formulas:
Horizontal:
Vx = Vxo
x = Vxt
Vertical:
y = 1/2 gt2 (on the cliff problems only)
Vy = Vyo - gt
y = Vyot - 1/2 gt2
Vy2 = Vyo2 - 2gy
non-horizontally launched projectiles
(7.2 Projectile Motion at an angle)
How can you predict the range of a launched marble?
Projectile Motion and the Velocity Vector

The path a projectile
follows is called its
trajectory.

The distance the
projectile travels is its
range.
Projectiles launched at an angle

A soccer ball kicked
off the ground is a
projectile, but it
starts with an initial
velocity that has
both vertical and
horizontal
components.
*The launch angle determines how the initial velocity divides
between vertical (y) and horizontal (x) directions.
Calculating the components
of a velocity vector
A soccer ball is kicked at a speed of 10
m/s and an angle of 30 degrees. Find
the horizontal and vertical components
of the ball’s initial velocity.
1.
2.
3.


Draw a free-body diagram
Use vx = v cos θ and vy = v sin θ.
Solve:
vx = (10 m/s)(cos 30o) = (10 m/s)(0.87) = 8.7 m/s
vy = (10 m/s)(sin 30o) = (10 m/s)(0.5) = 5 m/s
practice problem:
finding vector components
Find the component velocities
for a helicopter traveling 95
km/hr at an angle of 35° to the
ground.
95 km/hr
θ
vx = v(cos θ)
(95 km/hr)(cos 35) = 78 km/hr
vy = v(sin θ)
(95 km/hr)(sin 35) = 54 km/hr
practice problem:
Calculating acceleration on an incline
A submarine is moving at 10 m/s
toward the surface at an angle of
35° from the ocean floor. How long
will it take to surface if it is 250 m
deep?
Calculate the y component of the submarine’s
ascent:

vy = (10 m/s)(sin 35o) = 5.7 m/s
Calculate the time to ascend 250 m

?s/5.7 m X 250 m = 43.8 s

The horizontal distance a projectile moves can be
calculated according to the formula:

The vertical component of velocity alone is responsible for
giving the projectile its vertical height AND "hang" time.
y= -1/2gt2
"Hang Time"





You can easily calculate your own hang time.
Run toward a doorway and jump as high as you can, touching the wall
or door frame.
Have someone watch to see exactly how high you reach.
Measure this distance with a meter stick.
The vertical distance formula can be rearranged to solve for time:
Calculating range and velocity
Human powered
George Fox
Univ.
punkin chunkin trailer
world record air cannon
Upwardly Launched Projectiles
What’s missing?
What’s
staying
same?
the

What does gravity do to its horizontal speed?

NOTHING!!!!

But its vertical speed when it lands will equal the
speed at which at was launched.
Example of a Cannon
Which cat will go further?
A
B
C
Which cat will go higher?
A
B
C
Which cat will be in the air longer?
A
B
C
What determines how far the projectile
will go?
The initial angle of the projectile and its
horizontal velocity determine its range.
Projectiles launched at the same speed but
different angles.
What do you notice about the diagram?

A ball launched at a
steep angle will have a
large vertical velocity
component and a small
horizontal velocity.

A ball launched at a
low angle will have a
large horizontal
velocity component
and a small vertical
one.

Which angle produces the largest range for the
projectile?

45o

Which angle will
the
for the projectile?

90o
produce
longest hang time

The range of a projectile is calculated from the
horizontal velocity AND the time of flight.
If a ball is kicked at the same speed, angles that add up
to 90 degrees have the same range.
In those cases in which the vertical displacement is zero,
the horizontal displacement (also known as range) can be
expressed in terms of initial launch speed and angle:
Shot Put - 45º is not the ideal angle, why?
You are already off the ground 1-2 meters.
boy, have we got problems!!!
take some of these on
Be sure to have all these formulas:
Horizontal:
Vx = Vxo
x = Vxt
hang time
Vertical:
y = - 1/2 gt2 (on the cliff problems only)
Vy = Vyo - gt
y = Vyot - 1/2 gt2
Vy2 = Vyo2 - 2gy
range
An Australian football is kicked at an angle of
37.0° with a velocity of 20.0 m/s.
Find:
(a) the max height,
(b) the time in air before striking the ground,
(c) the horizontal distance traveled,
(d) the velocity vector at the max height, and
(e) the acceleration vector at max height.
Here’s what we know:
Displacement
Horizontal
?
Vertical
?
Time
Initial Velocity
?
Vo cos 37.0o
?
Vo sin 37.0o
?
?
?
?
0
9.8 m/s/s
Average
Velocity
Final Velocity
Acceleration
Where do you begin?
Look at the formulas again, they’re taunting you aren't they?
set-up the table and we can see we can find initial velocity for
both the horizontal and vertical:
Vxo = Vo cos 37.0o = (20.0 m/s)(0.799) = 16.0 m/s
Vyo = Vo sin 37.0o = (20.0 m/s)(0.602) = 12.0 m/s
fill in the table as we go along
(a)The max height is attained where Vy = 0,
this occurs when t = Vyo/a, ergo (12.0 m/s)/(9.80 m/s2) =
1.22 s
Time for another formula:
y = Vyot - 1/2 gt2
(12.0 m/s)(1.22 s) - 1/2(9.80 m/s2)(1.22 s)2
= 7.35 m.
=
(b) To find the time it takes for the ball to return to the
ground, we use the following equation and set y = 0 (for
ground level).
y = Vyot - 1/2 gt2
thus,
0 = (12.0 m/s)t - 1/2(9.80 m/s2)t2
t=
2(12. m/s)/
2
(9.80 m/s )
= 2.45 s
you might have found t = 0, this is a solution,
congratulations, but it is wrong, it is 0 when the initial point
y is zero.
(c) The total distance traveled horizontally is found by
applying the equation x = Vxot, remember: a = 0,
(remember from your table: Vxo = 16.0 m/s)
x = (16.0 m/s)(2.45 s) = 39.2 m
(d) The velocity at max height
At the max height there is no vertical component to the
velocity, only horizontal.
so v = Vxo cos 37.0o = 16.0 m/s.
(e) The acceleration vector at max height
The acceleration vector is always 9.80 m/s2 downward.
are you ready??
Test – Wednesday!!!
Calculating acceleration on an incline
A skier with a mass of 50 kg is on a
hill making an angle of 20 degrees.
The friction force is 30 N. What is
the skier’s acceleration?
1.
2.
3.



You know the mass, friction force, and angle.
Use relationships: a = Fnet ÷ m and Fx = mg sinθ.
Calculate the x component of the skier’s weight:
Fx = (50 kg)(9.8 m/s2) × (sin 20o) = 167.6 N
Calculate the force: Fnet = 167.6 N – 30 N = 137.6 N
Calculate the acceleration: a = 137.6 N ÷ 50 kg = 2.75 m/s2
Tuba Tony gets into a fight with his girlfriend Charity the
Cheer Queen at the last home football game of the year.
While Tuba Tony is on the field during the half time show,
Charity throws Tony’s class ring off the top of T-bird
Stadium.
.
The stadium is 10 m high and the ring was thrown at a
velocity of 10 m/s. Tony, distraught over the loss of his
class ring decides to search for it. Let’s use physics to
determine where he should look!

How long does it take for Tuba Tony’s ring to hit the
ground?

How far horizontally from the base of the stadium does
Tuba Tony’s class ring land?

How long does it take for Tuba Tony’s ring to hit
the ground?
y = ½ gt2
t = √ 2y/g
t = √2(10 m) / (10 m/s2)
t = 1.41 s

How far horizontally from the
base of the stadium does Tuba Tony’s
class ring land?
v=d/t
d=vxt
. d = 10 m/s x 1.41 s
. d = 14.1 m
Pirate monkeys on a boat are positioned at rest
100 m from the base of a 40 m high cliff. A
cannon sits on the cliff and fires horizontally at
the boat with a velocity of 35 m/s.

Will the cannonball hit the boat?
35 m/s
40 m
100 m

How long will it take the cannonball to hit the water?
d = ½ g t2
t = 2d / g
t = 2(40m) / (10m/s2)
t = 2.83 s

How far will it travel
from the base
v=d/t
d = 35 m/s x 2.83 s
= 99.05 m
horizontally
of the cliff?
.d=vxt
A mountain climber encounters a crevasse in an ice field.
The opposite side of the crevasse is 2.75 m lower and is
separated horizontally by a distance of 12 m. To cross the
crevasse, the climber gets a running start and takes off
horizontally. What speed will the climber have to take off
to make it safely cross?
Choose a formula:
y = – 1/2gt2
Solve for time:
-2.75 = -1/2gt2
t = 0.75 s
Calculate the horizontal velocity needed:
x = vxt
v = x/t
= 12/.75 = 16 m/s
A hunter aims his arrow directly at a target (on the same
level) 100 m away. If the arrow leaves the bow at a
speed of 75 m/s, by how much will it miss the target?
What should be the launch angle to hit the target?
1.3 s