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**Unformatted text preview: **50 m 3 - - - , § h | / Xo ball y on - unknown known - - - Vox Xo y Yo ball , = ( but Xf 50M Xfbau O = know we X = e- we you want ) ¥ .ir?io5aei=i.smis i
' O h = Yfbau I t d = you - O = Voy , I k i a. - ax G - = (4) Vo sin - - = 2.6ms = O , d ÷
DX Xo
We
in We know that order the time For , catch it to t VI - Vo ? = find want to how far out ball travels the 2a× ( Dx) t one use but X, uses of in the the X 2 kinematic direction , ✓ = Vo y
O and h t t t Z t Igt - gives x - Cvo since ) X - I will need the , so let's ±FEF4kgITh
Ya to that - - g be )
keep distance uses one doesn't help us plug this
+ into our X yza× O - to so , sin
- - - t = - 2.6 g - - t = end to up 3.475 9.8 . X Instead , . - xotvoxttyza.it let's try to find ' a isn't
time t
, , bad , but we using the y don't have direction displacement b" 4=0 xotvoxt 5.2 Im
want to know what Dff at Varg - - /vw = average velocity x d 3.47£
positive
Cvo
(6) EEE
t
= need we OH .sn/sC3.47s) We - 2C - us - = - * Za t where , ×=×o+vo×t
' Elad b ± - = Kay ' ( Vo sin 4) t solving for
t a×=o Now Yo t Voy t = that's
direction , because × that equations , the ball :
= the in . by trying to start might we we need in order to catch this ball . . I Note: checks still for Numbers made up bad this one . just fine ! work t
Ba = 2. Sanity are no here Unit good . known s - - - checks - - - - top - - Vo - - ¥7 III . . - - - . " '
- - - I = O fence ① OO
top of the At the V X s g - = 80 m 30 = 0 Xo
a) ai
- - - - - - , - - - - Vox 1 2 = - - Voy - Voy
now - c) If we can get the to peak , it will take Us to go up and down t Xf=l35.8mJ '
Y Yo tvg.tt Kay t equation displacement our use to Xf Ot 33.95 (4) 19.6 m/s = it takes zs Xf Xot Vox gL2s ) 2g = 4=0 , ay t t Voy - arc m ' ' - Y O - - t 19.6 K2) - Iz ( 9. 8) ( 4) 14--19.67
b) Now equation again but , ×= Xo + Vox t well Voy Vo X = = Now Y - - - (6) Vo sin = l!÷o . =
. Is 39.2 m/s = 33 95
. m/s Xot Vox t
t ( 33.95 )t 80 33795=2.36
plug Yo t in Voy 4=0+19.6/2.36 ) /#¥.96m s to ft
- Ya , t . the fence For this
. Let 's use the , we use can displacement our equation displacement in X . Kc ax =D 19.6 m/s = 19 Gm = Vo cos ( 30) 80=0
t a time Vox ? is , = need we height of tyzaxtz.IO Vo Sin ( 300) Vox the Vo×t X= Xot what find want to we ' ¥198712.365 d) Farag f) # at the peak Velocity is direction § in the only in the X so drag , - the
, X is direction . - ) a knowing
T T 0.03 = by 's 10 T slyr Wi
if b) R= 10km 1st we t = C- C- find to 2020 966 years = 3.05×10 let 's wot = 's IO Wo - - Wo = Wo 209.44 = Wo rads - 209.44 - t 276.32 get we to DI = wi-Wo = L l 1 year 209.37 = 209.44 - = A w 2.2×10-913.04×10 ) If = - 2.2×10-9 pulsar the 2.2×10-963.04×10 ) with constant 67 I w Wo = t stops
th - - is year rads s radha rotating , then t s - - ( 276.54 rads ) ( 10,000M ) If =C2.7Gsf
10000 |Ac=7.65xl08m/s€ - f 2.2×10-9 ) t 209.44 - - , rad Is
= Wf Lt 0=209.44 . = ac acceleration - 2.2×10-9 IJ-2.7654xiom.tt
t=3.O2x103y
V angular s find : 209.37rad Is = " WR = I year later so
, constant assume Dt t V slyr 0.03001 " t 209.44 rads = go.IT#o-sy=2T-rad/s = " 209.44 radls 31536000 q t to Lt }÷z = changes by A Years ' d use - t 1054 - = Now
w need 101000M = 107¥ - s increasing is Wo 9.53×1010 s o : 31536000 's ...

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- Fall '09
- Velocity, Trigraph, Want, ball