Angled Projectile Motion
2D Motion
2015
cjcb
RECAP: What is Projectile Motion?
 Projectile motion is the motion of an object that is being
thrown or launched.
 The object may be launched horizontally or at an angle.
 IN THE ABSENCE OF FRICTION OR AIR
RESISTANCE, the ONLY FORCE ACTING ON
A PROJECTILE IS THE FORCE OF
GRAVITY!!!! *****
(whether thrown straight up, horizontally, or at an
angle)
What Actually Happens?
http://www.physicsclassroom.com/class/vectors/u3l2e.cfm
What Does This Look Like?
KEY POINTS:
(1) HORIZONTAL and
VERTICAL motion are
treated separately
(2) HORIZONTAL motion is a
constant velocity
problem
(3) VERTICAL motion is like a
free fall problem
(acceleration, with a = g)
http://www.studyphysics.ca/newnotes/20/unit02_circulargravitation/chp07_2d/lesson26.htm
Solving Problems:
HORIZONTAL PROJECTILE MOTION
 Isolate the horizontal and vertical components into separate sides
and solve for each piece.
 Horizontal:
(constant velocity)
 vx=dx/t
• Vertical:
(free fall)
 dy = vyit + 1/2 gt2
 vyf2 = vyi2 + 2gdy
 vyf = vyi + gt
EXAMPLE: (Worksheet Prob #1 w/ extra parts)
Erica kicks a soccer ball at 12 m/s horizontally from the
edge of a roof of a building which is 30.0 m high.
QUESTIONS:
(a) WHEN does it strike the ground? (t = ? )
(b) How far from the building does it land? (dx = ? )
(added question)
(c) With what velocity does it strike the ground? (vxf & vyf)
STEP 1: DRAW PICTURE (down and out is POSITIVE)
STEP 2: IDENTIFY KNOWNS & UNKNOWNS
Picture:
HORIZ:
knowns:
vxi = 12 m/s
and also:
vxf = 12 m/s
VERT:
knowns:
dy = 30.0 m
(or dV)
vyi = 0 m/s
a = g = 9.8 m/s2
unknowns:
unknowns:
dx = ? (or dH)
vyf = ?
t=?
(time is same on both sides)
t=?
STEP 3: WRITE EQUATIONS – based on knowns
& unknowns & identify where have enough
information to solve for a variable
HORIZ:
knowns:
vi = vx = 12 m/s
EQUATIONS:
vx=dx/t
2 unknowns: t & dx
so can’t solve YET
VERT:
knowns:
dy = 30.0 m
vyi = 0 m/s
a = g = 9.8 m/s2
EQUATIONS:
dy = vyit + 1/2 gt2
only 1 unknown:
t
use this to SOLVE for time
(also know enough to solve for vyf in part c):
vyf2 = vyi2 + 2gdy
Erica kicks a soccer ball at 12 m/s horizontally from the edge of a roof of a building
which is 30.0 m high.
QUESTIONS:
(a) WHEN does it strike the ground? (t = ? )
 STEP 4: START SOLVING where have enough info so to do
 1st = TIME on VERTICAL SIDE
HORIZ:
knowns:
VERT:
knowns:
vi = 12 m/s
dy = 30.0 m
vyi = 0 m/s
a = g = 9.8 m/s2
EQUATIONS:
EQUATIONS:
vx=dx/t
(wait to work on this side)
dy = vyit + 1/2 gt2
since vyi=0 simplify and rewrite as:
t = (2 dy / g)1/2
t = (2 * 30.0 m) / 9.8 m/s2)1/2
t = 2.47 s
Erica kicks a soccer ball at 12 m/s horizontally from the edge of a roof of a building
which is 30.0 m high.
QUESTIONS:
(b) How far from the building does it land? (dx = ? )
Now NOTICE that TIME is in PURPLE. That’s because time is the SAME in the HORIZ. and
VERT. directions. SO, t calculated using equations on the vertical side can be used in
equations on the horizontal side and vice versa.
HORIZ:
knowns:
vi = 12 m/s
* Because constant velocity
vi = vx = vx= 12 m/s
EQUATIONS
vx=dx/t - rearrange:
dx = vx * t
dx = 12 m/s * 2.47 s
dx = 29.7 m from building
VERT:
we solved for:
t = 2.47 s
Erica kicks a soccer ball at 12 m/s horizontally from the edge of a roof of a building
which is 30.0 m high.
QUESTIONS:
(c) With what velocity does it strike the ground? (vxf & vyf )
 Now for part “c”. We KNOW the horizontal velocity when it lands because it’s the
same as the start (constant velocity). Use kinematic equations to solve for vyf .
HORIZ:
VERT:
Back to knowns & unknowns:
vxi = vxf = vx= 12 m/s
dy = 30.0 m
vyi = 0 m/s
a = g = 9.8 m/s2
t = 2.47 s
*solved for in part (a)
EQUATIONS:
OR
vyf2 = vyi2 + 2gdy
vyf = vyi + gt
(independent of t)
(using t from “part a”)
one way:
vyf = 0 m/s + (9.8 m/s2 * 2.47 s)
vyf = 24.3 m/s
Solving Problems:
ANGLED PROJECTILE MOTION
 NOW: What if you shot the cannon UP (or down) at
an angle instead of horizontally?
 It would look something like this:
Solving Problems:
ANGLED PROJECTILE MOTION
Angled Projectile Motion can START at the ground OR
“in the air” (at a different height than the final height)
But The Trajectory Is Always Parabolic!!
 symmetry of motion up and down
For ANGLED PROJECTILE MOTION:
• Still have constant vx, BUT it is no longer equal to
the overall initial velocity
• Still have a = g BUT no longer vyi = 0 m/s
-
only acceleration is “g”, in vertical direction
• There is an initial velocity in BOTH the
HORIZONTAL and VERTICAL directions
- but can calculate these two components
- just like last unit, use vector math (cos and sin) to resolve
velocity into vx and vyi
Solving Problems:
ANGLED PROJECTILE MOTION
 OUR NEW “STEP 1” is to break up the initial velocity into its “x”
(horizontal) and “y” (vertical) components:
 Horizontal:
(constant velocity)
vx = vi * cosq
• Vertical:
(constant acceleration)
vyi = vi * sinq
THEN Solve as for HORIZONTAL PROJECTILE
MOTION, using the vx and vyi calculated here.
Solving Problems:
ANGLED PROJECTILE MOTION
 EXAMPLE: Break up the initial velocity into its “x” (horizontal) and
“y” (vertical) components where vx = 50 m/s and q = 60° :
 Horizontal:
• Vertical:
(constant velocity)
(constant acceleration)
vx = vi * cosq
vx = 50 * cos60°
vx = 25 m/s
vyi = vi * sinq
vyi = 50 * sin60°
vyi = 43 m/s
Solving Problems:
ANGLED PROJECTILE MOTION
 For the rest of the problem, you STILL Isolate the horizontal and
vertical components into separate sides and solve for each piece.

SAME EQUATIONS as for HORIZONTAL
 Horizontal:
(constant velocity)
vx=dx/t
• Vertical:
 dy = vyit + 1/2 gt2
 vyf2 = vyi2 + 2gdy
 vyf = vyi + gt
Solving Problems:
ANGLED PROJECTILE MOTION
A BIT ABOUT “VOCABULARY”:
- “HANG TIME” refers to the time in the air (t)
-“RANGE” refers to the horizontal distance travelled (dx)
- “MAXIMUM HEIGHT” is the height where the
vertical velocity is zero (vy = 0 m/s)
Solving Problems:
ANGLED PROJECTILE MOTION
EXAMPLE: p. 152 #6
A rock is thrown from a 50.0 m high cliff with an initial
velocity of 7.0 m/s at an angle of 53.0° above the
horizontal. Find the velocity vector for when it hits the
ground. How long does it take?
 STEP 1: DRAW PICTURE (UP and out is POSITIVE)\
 STEP 2: Calculate the initial Velocity in the x (horizontal) and y
(vertical) directions, given vi = 7.0 m/s and angle of the projectile,
q = 53.0° (alternatively can substitute expressions for vx and vyi into
basic equations)
Picture:
Calculating initial velocity:
:
 Horizontal:
• Vertical:
(constant velocity)
(constant acceleration)
vx = vi * cosq
vyi = vi * sinq
vyi = 7.0 * sin53.0°
vyi = 5.59 m/s
vx = 7.0 * cos53.0°
vx = 4.21m/s
 STEP 3: IDENTIFY KNOWNS & UNKNOWNS
HORIZ:
knowns:
VERT:
knowns:
vx = vi * cosq = 4.21 m/s
unknowns:
vyi = vi * sinq = 5.59 m/s
dy = - 50.0 m
a = g = - 9.8 m/s2
unknowns:
dx = ? (or dH)
t=?
(time is same on both sides)
vyf = ?
t=?
** DOES THIS LOOK FAMILIAR? [See the
HORIZONTAL PROJECTILE MOTION example]
STEP 4: WRITE EQUATIONS – based on knowns & unknowns & identify
where have enough information to solve for a variable
HORIZ:
knowns:
vx
VERT:
knowns:
dy , vyi , and a = g
EQUATIONS:
EQUATIONS:
vx=dx/t
2 unknowns: t & dx
- this problem doesn’t ask for
t & dx, but you could solve for
them as in the earlier example of
horizontal motion (t from VERT
then substitute t into equation above)
vyf2 = vyi2 + 2gdy
only 1 unknown: vyf
dy = vyit + 1/2 gt2
only 1 unknown: t
- could use this to SOLVE for time
STEP 4a: SOLVE
FIRST: FIND EACH INDIVIDUAL COMPONENT OF FINAL VELOCITY:
HORIZ:
vxi = vxf = vx= 4.21 m/s
VERT:
vyf2 = vyi2 + 2gdy
(independent of t)
Substituting:
vyf2 = [vi * sinq] 2 + 2gdy = 5.59 m/s
vyf2 = [5.59 m/s]2 + 2(-9.8 m/s2)(-50.0 m)
vyf = 31.8 m/s
STEP 4b: SOLVE
SECOND: COMBINE INTO SINGLE VELOCITY
VECTOR – ADDING VECTORS AS IN THE LAST UNIT:
use Pythagorean Theorem
SO: Vf 2 = vx2 + vyf2
Vf 2 = (4.21 m/s)2 + (31.8 m/s)2
Vf = 32.08 m/s
AND
tan q = vyf / vx
tan q = 31.8 m/s/ 4.21 m/s
q = tan-1 (31.8 /4.21)
q = 82.45°
** We will also talk about solving for RANGE,
HANG TIME, and MAXIMUM HEIGHT **
Homework
 Angled Projectile Motion Worksheet, #1-3
 Page 152, #4-6