stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on its way down as shown in Figure 2.14. (A) Using l = 0 as the time the stone leaves the thrower's hand at position 0. determine the time at which the stone reaches its maxi- mum height. SOL LTIOL

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N Example 2.10
Not a Bad Throw for a Rookie!AM
V
Astone thrown from the top of a building is given an initial velocity
of 20.0 m/s straight upward. The stone is launched 50.0 m above the
ground, and the stone justI misses the edge of the roof on its way down
as shown in Figure 2.14.
t=
ye = 20.4 m
2.04 s
se--9.80 m/s
(A) Using la = 0 as the time the stone leaves the thrower's hand at
position O, determine the time at which the stone reaches its maxi-
mum height.
sa= 20.0 m/s
ae= -9.80 m/s
© le = 4.08 s
Ye= 0
e- -20.0 nm/s
e= -9.80 m/s?
SOLUTION
Ye
with dropping ohi-
nole 2.
) Posi n,
u OW-
veloe
aes ar
ing
"pwar and
a freely f ing ne
nes
p.
av
tity
Ma
ies
- ienc
10 simula
a the
sels r-
on
tin-
small e
er
I r val-
the
O la = 5.00 s
Ye- -22.5 m
",e= -29.0 m/s
4- -9.80 m/s2
sat
quir
50.0 m
bi-
of a' aildi
ie r
ston"
s ? art
der e
ion
*non
d
.oc is positi
wi chan si
i dal
ne ve c
tone
sint. F
1 ahvays
ar derati a of
have ner
e
= 5.83 s
0 = -50.0 m
",e= -37.1 m/s
Se= -9.80 m/s
de
nat it al a
e al
noose an
i ial
st after
- st
leaves 1e er
and a final
oint at th cop of i ight.
rime at whic
V = v. + a,1
Transcribed Image Text:N Example 2.10 Not a Bad Throw for a Rookie!AM V Astone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone justI misses the edge of the roof on its way down as shown in Figure 2.14. t= ye = 20.4 m 2.04 s se--9.80 m/s (A) Using la = 0 as the time the stone leaves the thrower's hand at position O, determine the time at which the stone reaches its maxi- mum height. sa= 20.0 m/s ae= -9.80 m/s © le = 4.08 s Ye= 0 e- -20.0 nm/s e= -9.80 m/s? SOLUTION Ye with dropping ohi- nole 2. ) Posi n, u OW- veloe aes ar ing "pwar and a freely f ing ne nes p. av tity Ma ies - ienc 10 simula a the sels r- on tin- small e er I r val- the O la = 5.00 s Ye- -22.5 m ",e= -29.0 m/s 4- -9.80 m/s2 sat quir 50.0 m bi- of a' aildi ie r ston" s ? art der e ion *non d .oc is positi wi chan si i dal ne ve c tone sint. F 1 ahvays ar derati a of have ner e = 5.83 s 0 = -50.0 m ",e= -37.1 m/s Se= -9.80 m/s de nat it al a e al noose an i ial st after - st leaves 1e er and a final oint at th cop of i ight. rime at whic V = v. + a,1
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