Lecture 04 A

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More Projectile Motion Discussion:Examples

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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


y# , 0()- m4 vi , $- ms4 3i , &-;


building where it lands$'irst, calculate

vxi , vi cos53i7 , .& ms

vyi , vi sin53i7 , -- ms a7 #ime to hit the ground$ (

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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


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


vy# , vyi 0 gt

y# , vyi t 0 567gt$

5vy# 7 $ , 5vyi7$ 0 $gy#

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y# , 0()- m4 vi , $- ms4 3i , &-;


building where it lands$'irst, calculate


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


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

= + − → = −

°

= = =

Lecture 04 A

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