Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
CHAPTER 3 NOTES
TWO DIMENSIONAL MOTION AND VECTORS
Sect. 3-2: Equations of Kinematics in 2-Dimensions
Equations representing motion of an object on the x-axis
 _________________________________________
 _________________________________________
 _________________________________________
 _________________________________________
Equations representing motion of an object on the y-axis
 _________________________________________
 _________________________________________
 _________________________________________
 _________________________________________

The ‘x’ part of the motion occurs exactly as it would if the ‘y’ part did not occur at all, and vice versa.
Sect. 3-3: Projectile Motion
What is a projectile?



Projectiles are objects ________________________________________________________________
_______________________. Free fall implies _____________________________________________
o ____________________________________________ ________________________________
dimensions.
o Examples of projectiles include
 ______________________________________________________________________
 ______________________________________________________________________
_________________________________________________________
_______________________________________________________________________. Trajectory is
__________________________________________________________________________________
__________________________________________________________________________________
o We can ______________________________________________________________________
 __________________________________________ and
 __________________________________________
If a ball is thrown horizontally and another ball is dropped-----they will both
o ____________________________________________!!!!!!!!
1
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
Solving Projectile Motion Problems

You need to _________________________________________________________________
o 1st part _______________________________________
o 2nd Part _______________________________________
o Keep the _____________________________________________________________
Horizontal motion equations and constants
Vertical motion equations and constants
Calculating time of fall


If a projectile is _____________________________________________________________________
_____________________________________________________________________
For such projectiles, ________________________________________________________________
_____________________________________________________________________
2
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
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.
o You will have to _______________________________________________________
____________________________________
X-Component
Y-Component
Step 2: Write down equations that might solve it.



We can use ______________________________________________________________
 Your Y-equation ________________________
We then use _____________________________________________________________
 Your X-equation _______________________
These are the two most common equations I will use in this section!!!!
Step 3: Carry out the solution.
Time for the jump
Step 4: Carry out the solution.
Using our time to calculate the initial horizontal velocity
3
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
Practice Problems Below: (We shall do the first one together)
#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?
#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 cats speed when it slid off
the table?
4
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
Physics – Chapter 3.3
Projectile Worksheet (H/W)
1. A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8 m
horizontally before it hits the water below. What is the pelican’s speed?
2. If the pelican in question 1 was traveling at the same speed but was only 2.7 m above the water,
how far would the fish travel horizontally before hitting the water below?
3. Determine the launch speed of a horizontally launched projectile that lands 26.3 meters from the
base of a 19.3-meter high cliff.
5
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
4. A soccer ball is kicked horizontally at 15.8 m/s off the top of a field house and lands 33.9 meters
from
the base of the field house. Determine the height of the field house.
5. A ball is projected horizontally from the top of a 92.0-meter high cliff with an initial speed of 19.8
m/s. Determine: (a) the horizontal displacement,
(b) the final speed the instant prior to hitting the ground.
6
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
Projectiles launched at an angle:


As shown here, projectiles launched at an angle have ________________________________
______________________________________________
Suppose the initial velocity vector makes an angle θ with the horizontal
The ___________________________________ can be used to find ____________________
____________________________________________________
o _________________________
o _________________________
Practice: The Kickoff

A placekicker kicks a football at an angle of 40.0 degrees and the initial speed of the ball is 22 m/s.
Ignoring air resistance, determine the following:
a) Maximum height that the ball attains.
b) Time of flight between kickoff and landing
c) The range ‘R’ of the projectile
o How is this different from the problems you’ve solved so far?
_________________________________________________________________________
o How will we get horizontal and vertical velocity?
_________________________________________________________________________
Problem Sketch
7
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
Conceptual Example: Two Ways to Throw a Stone
From the top of a cliff, a person throws two stones. The stones have identical initial speeds, but stone 1
is thrown downward at some angle above the horizontal and stone 2 is thrown at the same angle below
the horizontal. Neglecting air resistance, which stone, if either, strikes the water with greater velocity?
Explain 
8
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
Projectiles launched at an angle worksheet - I
Show your work for full credit!!
1. An object is launched with an initial velocity of 25.0 m/s at an angle of 30° above horizontal.
a. Determine the components of the initial velocity. [answers: vox = 21.7 m/s, voy = 12.5 m/s]
b. Use the correct component to determine the time that it takes the object to reach the top of its
flight. [answers:1.27 s]
c. Calculate the maximum vertical height that the object will reach. [answers:7.96 m]
d. Calculate the horizontal distance that the object will travel. [answers:55.1 m]
9
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
2. A ball is launched with an initial horizontal velocity of 10.0 m/s. It takes 0.5 s for the ball to reach its
maximum height.
a. Determine the maximum horizontal distance that the ball will travel. [answers:10 m]
b. Calculate the initial vertical velocity of the ball. [answers:4.91 m/s]
3. If the initial velocity, equals 50.00 m/s, find the maximum height and range for each of the launch angles listed in the
table below.
10
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
Projectiles launched at an angle worksheet - II
Show your work for full credit!!
4. Measurements made in 1910 indicate that the common flea is an impressive jumper, given its size. Assume that
a flea’s initial speed is 2.2 m/s, and that it leaps at an angle of 21° with respect to the horizontal. If the jump
lasts 0.16 s,
a. What is the magnitude of the flea’s horizontal displacement?
b. How high does the flea jump?
5. A volleyball is spiked so that it has an initial velocity of
directed downward at an angle of
below
the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?
6. As a tennis ball is struck, it departs from the racket horizontally with a speed of
. The ball hits the
court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves
the racket? (notice that this is a horizontally launched projectile)
11
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
7. A skateboarder shoots off a ramp with a velocity of
, directed at an angle of
above the horizontal.
The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the
direction be
vertically upward, and take as the origin the point on the ground directly below the top of the ramp.
a. How high above the ground is the highest point that the skateboarder reaches?
b. When the skateboarder reaches the highest point, how far is this point horizontally from the end of the
ramp?
8. A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club
at a speed of
at an angle of
above the horizontal. It rises to its maximum height and then falls down
to the green. Ignoring air resistance, find the speed of the ball just before it lands.
12
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
KEY
5.
SSM REASONING The vertical component of the ball’s velocity v0 changes as the ball
approaches the opposing player. It changes due to the acceleration of gravity. However, the horizontal
component does not change, assuming that air resistance can be neglected. Hence, the horizontal
component of the ball’s velocity when the opposing player fields the ball is the same as it was
initially.
SOLUTION Using trigonometry, we find that the horizontal component is
b
g
v x  v 0 cos  15 m / s cos 55  8.6 m / s
6. REASONING
y  v0 y t 
We will determine the ball’s displacement y in the vertical direction from
1
a t2
2 y
(Equation 3.5b). We choose this equation because we know that v0 y  0.00 m / s
(the ball initially travels horizontally) and that a y  9.80 m/s2 (assuming that upward is the +y
direction). We do not know the value of the ball’s time of flight t. However, we can determine it
from the motion in the horizontal or x direction.
SOLUTION From Equation 3.5b we have that
y  v0 y t  a y t 2   0.00 m / s  t  a y t 2  a y t 2
1
2
1
2
1
2
The motion in the horizontal direction occurs at a constant velocity of v0 x  28.0 m / s and the
displacement in the horizontal direction is x  19.6 m . Thus, it follows that
x  v0 xt
or
t
x
19.6 m

 0.700 s
v0 x 28.0 m / s
Using this value for the time in the expression for y shows that
1
2
y  a yt 2 
1
2
 9.80 m/s2   0.700 s 2  2.40 m
Since the displacement of the ball is 2.40 m downward, the ball is 2.40 m above the court when it
leaves the racket.
13
Name: ___________________________ Period: _______________
7.
Date: __________
AKIBOLA
REASONING
a. Here is a summary of the data for the skateboarder, using v0 = 6.6 m/s and  = 58°:
y-Direction Data
y
?
ay
−9.80 m/s
2
vy
v0y
0 m/s
+(6.6 m/s)sin 58° = +5.6 m/s
t
x-Direction Data
x
?
ax
vx
v0x
0 m/s2
t
+(6.6 m/s)cos 58° = +3.5 m/s
We will use the relation v2y  v02y  2a y y (Equation 3.6b) to find the skateboarder’s vertical
displacement y above the end of the ramp. When the skateboarder is at the highest point, his vertical
displacement y1 above the ground is equal to his initial height y0 plus his vertical displacement y
above the end of the ramp: y1 = y0 + y. Next we will use v y  v0 y  a y t (Equation 3.3b) to calculate
the time t from the launch to the highest point, which, with x  v0 xt  12 axt 2 (Equation 3.5a), will give
us his horizontal displacement x.
SOLUTION
a. Substituting vy = 0 m/s into Equation 3.6b and solving for y, we find
 0 m/s   v02y  2a y y
2
or
y
v02 y
2a y

Therefore, the highest point y1 reached by
y1 = y0 + y = 1.2 m + 1.6 m = +2.8 m above the ground.
  5.6 m/s 

2
2 9.80 m/s2
the

 1.6 m
skateboarder
occurs
when
b. We now turn to the horizontal displacement, which requires that we first find the elapsed time t.
Putting vy = 0 m/s into Equation 3.3b and solving for t yields
v0 y
(1)
0 m/s  v0 y  a y t or t 
ay
Given that ax = 0 m/s2, the relation x  v0 xt  12 axt 2 (Equation 3.5a) reduces to x  v0 x t . Substituting
Equation (1) for t then gives the skateboarder’s horizontal displacement:
 v0 y  v0 x v0 y   3.5 m/s  5.6 m/s 
x  v0 xt  v0 x 

 2.0 m

 ay 
ay
9.80 m/s


14
Name: ___________________________ Period: _______________
Date: __________
AKIBOLA
8. REASONING The magnitude (or speed) v of the ball’s velocity is related to its x and y velocity
components
(vx and vy )
by
the
Pythagorean
theorem;
v  v x2  v 2y
(Equation 1.7).
The horizontal component vx of the ball’s velocity never changes during the flight, since, in the
absence of air resistance, there is no acceleration in the x direction (ax = 0 m/s2). Thus, vx is equal to
the horizontal component v0x of the initial velocity, or vx  v0 x  v0 cos 40.0 . Since v0 is known, vx
can be determined.
The vertical component vy of the ball’s velocity does change during the flight, since the acceleration
in the y direction is that due to gravity (ay = 9.80 m/s2). The relation v 2y  v02 y  2a y y (Equation
2.6b)
may
be
used
to
find
v 2y ,
since
ay,
y,
and
v0y
are
known
(v0y = v0 sin 40.0º).
SOLUTION The speed v of the golf ball just before it lands is
v  v x2  v 2y 
 v0 cos 40.0
2
 v02 y  2a y y
v 2y

 v0 cos 40.0   v0 sin 40.0
2
2
 2a y y


 14.0 m/s  cos 40.0   14.0 m/s  sin 40.0  2 9.80 m/s 2  3.00 m   11.7 m/s
2
2
15