Vertical Projectile Motion

1. Projectile Motion
With An Initial Vertical
Velocity Component
Includes symmetrical trajectory equations.Includes symmetrical trajectory equations.
Nelson Reference:Nelson Reference:
Page 39 (SP#2) to 42Page 39 (SP#2) to 42

2.  If an arrow, or any other object, is shot with anIf an arrow, or any other object, is shot with an
angle between 0angle between 000
and 90and 9000
to the horizontal, theto the horizontal, the
arrow (or object) will have an initial verticalarrow (or object) will have an initial vertical
velocity component. For these types of problems,velocity component. For these types of problems,
we must use the initial vertical velocitywe must use the initial vertical velocity
component (vcomponent (v11sinsinθ)θ)..
v H o r i z o n t a l = v 1 c o s ( )
vVertical=v1sin()
v 1
U p

3. Case 1: An arrow is shot from ground level and: An arrow is shot from ground level and
returns to ground level. This is areturns to ground level. This is a symmetrical
trajectory (ST), all green equations are for ST.
 Step 1: Find the time of flightStep 1: Find the time of flight
Δd vertical = v1 sin(θ)Δt + ½ a(Δt)2
, but, but Δd vertical = 0,,
andand a = - g
SoSo, Δt ( v1 sin(θ)- ½ g Δt) = 0,, after factoring out Δtafter factoring out Δt
Thus, Δt = 0 orThus, Δt = 0 or Δt = (2v1 sinθ)/g,, we ignore Δt = 0
G r o u n d L e v e l
U p
R a n g e
v
1

4.  We can now calculate the range:We can now calculate the range:
R =R = ((vv11 cosθ)cosθ)Δt =Δt = ((vv11 cosθ)cosθ)(2v(2v11 sinθ)/gsinθ)/g
R =R = (2(2vv11
22
cosθsinθ)/gcosθsinθ)/g, but, but 2sinθ·cosθ = sin2θ
So, R = (v1
2
sin(2θ)) / g, from this equation we can seefrom this equation we can see
that the greatest range is achieved whenthat the greatest range is achieved when θ = ____ ..
 To find the maximum height, we would use:To find the maximum height, we would use:
 Δt = 0.5·(2v1 sinθ)/g= (v1 sinθ)/g, this value is just half of
the total time of flight (symmetrical trajectory only)
 Δdmax-vertical =,( v1 sinθ)/g )·[(v1 sinθ)x ½],{Δd = vav x Δt }
 The square brackets represents the average vertical velocityThe square brackets represents the average vertical velocity
from ground level to the highest point reachedfrom ground level to the highest point reached
(where v(where vverticalvertical = 0)= 0)
 Δdmax-vertical = ( v1
2
sin2
θ)/(2g )

5. Case 2 – Not a Symmetrical Trajectory– Not a Symmetrical Trajectory
 In the diagram, we see that the arrow achieves a finalIn the diagram, we see that the arrow achieves a final
height of “h”. When we calculate the time of flight, weheight of “h”. When we calculate the time of flight, we
would get two different and correct values.would get two different and correct values. OneOne valuevalue
would represent the arrow reaching height “h” on the waywould represent the arrow reaching height “h” on the way
up ( at “A”). Theup ( at “A”). The secondsecond time value occurs when thetime value occurs when the
arrow reaches the same height on the way down (at “B”).arrow reaches the same height on the way down (at “B”).
To know which value is correct, more information wouldTo know which value is correct, more information would
have to be given (range, achieving max height, etc.).have to be given (range, achieving max height, etc.).
 To find time of flight, in this example, a quadratic equationTo find time of flight, in this example, a quadratic equation
would have to be solved since hwould have to be solved since h ≠≠ 00
((h = 0 only for a symmetrical trajectoryh = 0 only for a symmetrical trajectory).).
R a n g e
v1
h
A B

6. Chalkboard ExampleChalkboard Example
 An arrow is shot with an initial velocity ofAn arrow is shot with an initial velocity of 10.0 m/s10.0 m/s
[30[3000
Up from Horiz.]Up from Horiz.] from a bridge which isfrom a bridge which is 10.0 m10.0 m
above the ground. Lettingabove the ground. Letting g = 10.0 m/sg = 10.0 m/s22
,,
determine the following:determine the following:
1.1. Time of flightTime of flight
2.2. Maximum height above the groundMaximum height above the ground
3.3. RangeRange [Ans. 2.0 s, h[Ans. 2.0 s, hmaxmax= 11.2 m, R = 17.4 m]= 11.2 m, R = 17.4 m]
G r o u n d L e v e l
v 1
R a n g eh
U p
h M a x

7. Practice Problems and VideosPractice Problems and Videos

http://www.youtube.com/watch?v=ZBfy-MNgtoYhttp://www.youtube.com/watch?v=ZBfy-MNgtoY (Very short)(Very short)

http://www.youtube.com/watch?v=cxvsHNRXLjwhttp://www.youtube.com/watch?v=cxvsHNRXLjw
(Shooting at a toy monkey which drops from a tree(Shooting at a toy monkey which drops from a tree
– where do you aim?)– where do you aim?)

http://www.youtube.com/watch?v=-uUsUaPJUc0http://www.youtube.com/watch?v=-uUsUaPJUc0
(Long video – not shown in class)(Long video – not shown in class)
 Practice questionsPractice questions copied from Physics 12,copied from Physics 12,
McGraw-Hill. Same handout as used in firstMcGraw-Hill. Same handout as used in first
class on PM.class on PM.
 Physics Rap!Physics Rap!
http://www.youtube.com/watch?v=3X5-nsbPz88http://www.youtube.com/watch?v=3X5-nsbPz88

8. Practice Questions
 Nelson: Page 40 # 4, Page 43 #2, 4, 6Nelson: Page 40 # 4, Page 43 #2, 4, 6
 Workbook page 17 Q # 1, 5, 7Workbook page 17 Q # 1, 5, 7
 Workbook page 18 Q # 4, 6, 7Workbook page 18 Q # 4, 6, 7
 Symmetrical Trajectory Questions:Symmetrical Trajectory Questions:
 Workbook page 18 Q # 1Workbook page 18 Q # 1
 Nelson Page 42 #1-3Nelson Page 42 #1-3