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33.144 we saw: 214" Q 1‘“ a) With what speed must a ball be thrown vertically from B O chaur‘r’ BE am, 9:, UéE ; _ ___.. ... ‘—‘—r gram level :0 rise to a maximum height of 50 m? (b) How long y =50 ”4
willitbeinlhe_air?(c)8kc3chgraphsofy,v,andaversusrfm" V. _ M the ball. 0n the ﬁrst two graphs, indicate the time at which so In  O “‘5 is reached. am .. ., a .. .. . 3 “£5. y' l ﬂl wig burgh! 70 £35.31; 2215 JwIT/AL ”EMT: V; ganja 71%: 49:93:77th THAT THE FIAaﬁL— 7 sz» THE TIME 774.9: 31414.. SPENbS 1A5 1.5 ML (4.95 cm 05;:
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A3035 MULTIPLJ)’ BY 2: 2—“ ____i_...q....t__' . __ HEW 6 52A stone is dropped into a river from a bridge 43.9 m above _ '.
the water. Another stone is tluown Venically down LOO s after the i
ﬁrst is dropped. Both stones strike the wane: at the same time.
(a) What is the initial speed of the second stone? (1)) Pint velocity
vetsus time on a graph for each stone, taking zen: time as the instant
the ﬁrst stone is released. PM?” 2224“: A PIWRE # Loam—E A' 25? szl I must?) Peon mimosa mint 33.3.1: ,
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EEE i .x. (a A car moving with cmslant nucleation covered the distance GIUEM o
between two points 60.0 m apattiin 6m 5. Its speed as it passes'
the second point was 15.0 mls. {ail W'hal was the speed at the ﬁrst
point? ('3) What was the acceleration? (c) At what prior distance
from the ﬁrm pair“ was ihe car at test? ((1) Graph 1 versus 1' and v
Ivmusrforﬂlecax. from rest (r = 0). um um: 1924;,— MwA PIJURE + LoamE A “RyT Pkwy,
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22144 200 SHIHS 22—!“ CHZ‘x',' ' HKWS [OSA pamchutisl bails out and freely falls 50 111. Then the pan»
chute opens. and thereafter she deceleratcs at 2.0 mfsz. She reaches
the ground with a speed 013.0 mix. (a) How long is the parachukist in the air? (b) {M what height (10:5 the fall begin? a B “'C = + 2:53:31
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311“ 2003"!” . 22141 @_ Cal{3' a‘=(3.0m)i+(4.0m)j aml b=(5.0m)i+( 2.0m)j give 5’ + b m (a) unit vegtot mttation an_d as (b) a magnitude ant
(c) an angle (relative to E). New give {3 — a in (d) unitvector
notation and as (e) a magnihtde and (f) an angle '41; Forthcvectors  _ I  HKWS I h I
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H.111: tWDIVectors E” and b in Fig. 329 have equal magnitudes . i'
of 10.0 In. Find (a) the 1 component and (b) the ) component of I ' N OTE ' [as  £90 3 S E I their vector sum F'. (c) the mag J, .hitude of F. and (d) the angle F‘
makes with the positive direction
of the: axis. sun ttw m Fig. 529 Problem 2]. 50 SHEETS at =§ _  _ _J_ _.
§§ FIR???" F’mb 7mg COMPDMEMt—S fall a AND bf.
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22144 200 SHEITS 22!“ ® _ 3‘ J Use the deﬁnition of scalar product, E 5. = ab :05 8. and the
fact that E F = all), + ayb’; + art): [see Exercise 30).“: Calculale
thc_angle Eetwetq the My \rcctors given by a‘ = 3.0i ~I_ 3.0j +
3.0:: and b = 2.0: + 1.0, + mi. ssrn nu ma: ﬂ .
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rim of the wheel is at the point of .. ppm .912. MONITUhE
contact hettimcn the wheel and ; AN}: mIEWATquﬁm.
the ﬂoor. At. a later time :2. the 3.5.1:. ;...... ._. __ ___ . __ wheel has rolled throughbnehalf At time ,1 A. time ,2 _ EH. K "' q 5‘ O "' M
of a revolution. What are (a) the _ ' ' _.._. 0, ”50 M
magnitude and (b) the angle (set R" 5 H Problem 7' alive to the ﬂoor) orthetlisplaccment odeuIing this interval? sum t—mrE' bLoaATEAﬁEJTFW‘ X l  . ﬁﬁmﬁz
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magnitude 6.0 units differ in directions by 60'. Find (a) the scalar 
product of Ihgtwu Vectoe's' and (b) the magnitude of the vector
product i” x b. 55th E’T: = ﬂame! Com m '60") 30
EEEEE = 003(Q ALW‘CGD") I
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22!“ 200 SHIITS 211 41 [1:] m1. Ashipselsuultosailtoapoiml20kmduenorth.An GWEM 6L. = [2.0km
unexpected storm blows the ship to a point 100 km due east of its _ l: I
starting point. How fax. and in what dinaction. must it now sail to b "' I ‘90 reach its original destination? k
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a 5— 145‘ “E. fFor'ihe mots in Fig. 3.35. calculate(a) n b. (b) 31:, and
(c) hoe.  ' .F' M” FIND “5:. I3:
I . —. _ Z _ d o‘ 1: l q 30
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 Spring '08
 Ramsdell/Wick
 Physics

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