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8/18/2019 Lecture 04 A
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More Projectile Motion Discussion:Examples
8/18/2019 Lecture 04 A
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More Projectile Motion Discussion:Examples
I hope this doesn’t
apply to you!
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Solving Projectile Motion Problems
"ead the problem care#ully, & choose the object(s) youare going to analyze.
$ S%etch a diagram.
& 'hoose an origin & a coordinate system.
( Decide on the time interval; this is the same in both
directions, & includes only the time the object ismoving with constant acceleration g
) Solve for the x and y motions separately.
* +ist known & unknown uantities. !emember that vx never changes, & that vy , - at the highest point.
. Plan how you will proceed. "se the appropriate
euations; you may have to combine some of them.
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Example ((: /on0Symmetric Projectile Motion
1inematic E2uations
vxi , vicos3i4 vyi , visin3ivx# , vxi 4 x# , vxi t
vy# , vyi 0 gt
y# , vyi t 0 567gt$
5vy# 7 $ , 5vyi7$ 0 $gy#
8 stone is thro9n! xi , yi , -
y# , 0()- m4 vi , $- ms4 3i , &-;
a7 #ime to hit the ground$b7 %peed just before it hits$c7 istance from the base of the
building where it lands$
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Example ((: Solution
8 stone is thro9n! xi , yi , -
y# , 0()- m4 vi , $- ms4 3i , &-;
a7 #ime to hit the ground$b7 %peed just before it hits$c7 istance from the base of the
building where it lands$'irst, calculate
vxi , vi cos53i7 , .& ms
vyi , vi sin53i7 , -- ms a7 #ime to hit the ground$ (
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Example ((: Solution
8 stone is thro9n! xi , yi , -
y# , 0()- m4 vi , $- ms4 3i , &-;
a7 #ime to hit the ground$b7 %peed just before it hits$c7 istance from the base of the building
where it lands$'irst, calculate
vxi , vi cos53i7 , .& ms
vyi , vi sin53i7 , -- ms thit , ($$ s
b7 +elocity just before it hits$vx# , vxi 4 vy# , vyi = gt so vx# , .& ms
vy# , - = 5>?75($$7 , 0 && ms
%peed 5v# 7$ , 5vx# 7$ @ 5vy# 7$
v# , &)? msngle* tan53# 7 , 5vy# vx# 7 , 05&&.&7 , 0?
3# , 0*->;
1inematic E2uations
vxi , vicos3i4 vyi , visin3ivx# , vxi 4 x# , vxi t
vy# , vyi 0 gt
y# , vyi t 0 567gt$
5vy# 7 $ , 5vyi7$ 0 $gy#
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Example ((: Solution
8 stone is thro9n! xi , yi , -
y# , 0()- m4 vi , $- ms4 3i , &-;
a7 #ime to hit the ground$b7 %peed just before it hits$c7 istance from the base of the
building where it lands$'irst, calculate
vxi , vi cos53i7 , .& ms
vyi , vi sin53i7 , -- ms thit , ($$ s
v# , &)? ms4 3# , 0*->;
c7 istance from the base of th building where it lands$
x# , vxi thit , 5.&75($$7 , .&- m
1inematic E2uations
vxi , vicos3i4 vyi , visin3ivx# , vxi 4 x# , vxi t
vy# , vyi 0 gt
y# , vyi t 0 567gt$
5vy# 7 $ , 5vyi7$ 0 $gy#
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Example ($:
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( m
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longjumper leaves the ground at an angle 3i , $-B above thehorizontal at a speed of vi , ?- ms.
a7 -ow far does he jump in the horizontal direction$ (ssume his motion is euivalent to that of a particle.) b7 hat is the ma/imum height reached$
1inematic E2uationsvxi , vicos3i4 vyi , visin3i 4vx# , vxi
x# , vxi t4 vy# , vyi = gt
y# , vyi t 0 567gt$
5vy# 7 $ , 5vyi7$ 0 $gy#
vxi , vi cos53i7 , .) ms
vyi
, vi
sin53i
7 , (- ms
" , .>( m b7 hat is the ma/imum height$
h , C5vyi7$5$g7
h , -.$ m
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Example: Driving ## a 'li##!!
vx# , vxi , F vy# , 0gt
x# , vx# t4 y# , 0 567gt$
svx- , 5xt7 , $?$ ms
movie stunt driver on a motorcycle speeds horizontally off a )--0mhighcliff. -ow fast must the motorcycle leave the cliff top to land on levelground below, >-- m from the base of the cliff where the cameras are$
1inematic E2uations: vxi , vicos3i4 vyi , visin3i 4vx# , vxi x# , vxi t
vy# , vyi = gt4 y# , vyi t 0 567gt$4 5vy# 7
$ , 5vyi7$ 0 $gy#
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Solutions: Driving ## a 'li##!!
vx# , vxi , F vy# , 0gt
x# , vx# t4 y# , 0 567gt$
svx- , 5xt7 , $?$ ms
movie stunt driver on a motorcycle speeds horizontally off a )--0mhighcliff. -ow fast must the motorcycle leave the cliff top to land on levelground below, >-- m from the base of the cliff where the cameras are$
1inematic E2uations: vxi , vicos3i4 vyi , visin3i 4vx# , vxi x# , vxi t
vy# , vyi = gt4 y# , vyi t 0 567gt$4 5vy# 7
$ , 5vyi7$ 0 $gy#
vx , vxi , F4 vy# , 0gt
x# , vxit4 y# , 0 567gt$
#ime to the bottom 0time when y , 0 )- m
0 567gt$ , 0 )- m
t , &> st that time x# , >-- m%o vxi , 5x# t7 , 5>-&>7
vxi , $?$ ms
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Example: 1ic%ed Hootball
2 football is kicked at an angle 3- , &.-B with a velocity of$-- ms, as shown. 'alculate:a #he ma/imum height. b #he time when it hits the ground.c #he total distance traveled in the x direction.d #he velocity at the top. e #he acceleration at the top.
3- , &.;4 v- , $- ms
vx-, v-cos53-7 , * ms4 vy-, v-sin53-7 , $ ms
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'onceptual Example
Demonstration!!
vx-
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'onceptual Example: rong Strategy
JShooting the Mon%eyK!!
Lideo 'lips!!
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Example: 8 Punt!
vi , $- ms4 3i , &.;
vxi , vicos53i7 , * ms4 vyi, visin53i7 , $ ms
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Proof that the projectile path is a parabola
x# , vxi t 4 y# , vyi t = 567g t$
/ote: #he same time t enters both euations⇒ 3liminate t to get y as a function of x.
%olve the x euation for t* t , x# vxi
4et* y# , vyi 5x# vxi7 = 567g 5x# vxi7$
5r* y# , 5vyi vxi7x# 0 C567g5vxi7$5x# 7$
#his is of the form y# , 8x# = G5x# 7$
A parabola in the x-y plane!!
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Example :
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Problem
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Problem Solution7hoose the origin at ground level, under where the projectile is launched, &up to be the positive y direction. 'or the projectile*
a #he time to reach the ground is found from the free fall euation, withfinal height 0 -. 7hoose positive time since the projectile was launched at t , -.
b #he horizontal range is found from the horizontal motion atconstant velocity.
89:.8m s ,v = 8 ;:.8 ,θ = ° , ya g = − 8 s , =.;9::s ?.?9s
=
y y y y v t a t y v t gt
v v g t
g
y
θ
θ θ
= + + → = + − →
− ± − −= = − =
−
( ) ( ) ( ) ( )8 8cos 9:.8 m s cos :.8 ?.?9>s :
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c t the instant just before the particle reaches the ground, thehorizontal component of its velocity is the constant
#he vertical component of velocity is found from*
d #he magnitude of the velocity is found from the x and y components calculated in part c above.
( ) ( ) ( )=8 8 8sin 9:.8 m s sin ;:.8 ?.@8 m s ?.?9>s
98.>m s
y yv v at v gt θ = + = − = ° −
= −
( )8 8cos 9:.8 m s cos:.8 :.= m s . xv v θ = = ° =
( ) ( )= == = :;.= m s 98.> m s @8.:m s x yv v v= + = + − =
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e #he direction of the velocity is
so the object is moving
# #he ma/imum height above the cliff top reached by the projectile
will occur when the y0velocity is -, and is found from*
< < 98.>tan tan >@.9
:;.=
y
x
v
vθ
− − −
= = = − °
>@.9 below the horizon .°
( )
( )
( )
= = = =
8 8 8 8 ma/
= == =
8 8
ma/ =
= 8 sin =
9:.8 m s sin ;:.8sin
A8.?m= = ?.@8 m s
y y yv v a y y v gy
v
y g
θ
θ
= + − → = −
°
= = =