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PROJECTILE MOTION SPH4U – Unit 1 Dynamics (pg. 36-43)
Transcript of PROJECTILE MOTION - Mr.Neave's Websitemrneave.weebly.com/uploads/1/3/5/9/13590915/4u1...SOLVING...
PROJECTILE MOTION SPH4U – Unit 1 Dynamics (pg. 36-43)
PROJECTILE MOTION
| The motion of a projectile such that the horizontal component of the velocity is constant, and the vertical motion has a constant acceleration due to gravity
PROPERTIES OF PROJECTILE MOTION
| The horizontal motion of a projectile is uniform | The horizontal component of acceleration of a
projectile is zero | The vertical acceleration of a projectile is constant
because of gravity | The horizontal and vertical motions of a projectile
are independent, but they share the same time!!
ANALYZING PROJECTILE MOTION
KINEMATICS EQUATIONS
SOLVING PROJECTILE MOTION PROBLEMS
| thorizontal = tvertical
| Break the motion into its horizontal and vertical components
| Use the appropriate kinematic equation(s) y apply constant velocity equations for horizontal
displacement y apply constant acceleration equations for vertical motion
a o
a=g= 9.845ft ]
TYPES OF PROJECTILE MOTION PROBLEMS
| Launching a projectile horizontally
| Launching a projectile at an angle y Lands at same level as launch (special case) y Lands at a different level Usually the best strategy is to find the time of flight first !!
PROJECTILE LANDS AT SAME LEVEL AS LAUNCH
| Range – the horizontal displacement of a projectile
EXAMPLE QUESTIONS Pg . 40 #4
e##.ES#a3i3HY:*I Ddx 1
.
x yddx= ?
a×=Oys2 Ddy= - 2.sn
ay=- 9.8%2
vi ,n=43cos42°viy ,-4.3 sink
at = ? dt= ?
Required at1¥ →tT+
Analysisi_ddyivjottlaat2Idx-vixat@aEt2steps_bdy.v
: yattgadt'
- 2.5 - = -4.3 ;dsn4Itttf9E⇒at'
- 2 .sn = -2.877% At -49}
ata b c
49¥ at't 2.877g at -2 .5n=O
at .
- b±F÷Za
at = . 2.877±M28y?#÷9.8
at .la?dt68at=28zl#681st = 0.478 s at = can't have
= 4.3nA ( cos 42'X0.4784
xd× =1 . Sm ✓
Speed when ball is caught
¥×= ?Vty= ?
Dt= 0.4785 At . 0.478 s
AdX= 1.53in Ady= -25in
k×=0mµ 8y= -9845Vix .43§as 42
'
Vijt .
' 4.3 ;ssh42°=3 .196M/s
=Z877m/sVtxikxtnmifof,⇒ Y=vitaIY×=3t96nf
¥-28749.8;D
. 1964: lots Yyitahetfefgffn :p
lifted Vajtsbmf14=8.245
OBJECT LANDS AT A DIFFERENT HEIGHT THAN LAUNCH HEIGHT
(Pg .43
D- ddx
ddfsd >Blsdx-150-185
soprojeotikdoesit bdyTl3mhitsideofbulding
Giveniy.45%645) Vix '4§fioi)
bdy - Bm axiomsay=
- 98%2 atx . ?bty: ? btytstx
Required: adx
Analysis: Find At when ddy=Bm
Check that projectile clears
side ofbuilding .
Steps rsdy :vjsinotttlgaat
13 m , 45mg sin 35ft ¥9.8 ;D at
13ns. 25.8 At -4.9 At '
4.9 at2- 25.81T +13 = O
At = - b tub2 a
HOMEWORK
| Pg. 40 Practice #1-3 | Pg. 42 Practice #1-3 | Pg. 43 #4
✓
0.
**g40*1-342*1
Pg . 40*1Horizontal 0°
F.. .
.
Given.
-0=004g=0ysVix =/
. 93mF
ay . -9.8%2ajomlss
Ddy= - 0.765mi
atiatg IHI !
*
;qo÷:::iIi←nontauy =O°
-
bdy .
:O. 83N
adx = 18.4 m
oh, =0mKay =
-9.9mL
;at . ? at×=otyViy=0m/s
