to that. The corresponding distance is
y
–
y'
= 0.50
h
, where
y
denotes the location of the
ground. In these terms,
y
is the same as
h
, so we have
h
–
y'
= 0.50
h
or 0.50
h
=
y'
.
(a) We find
t'
and
t
from Eq. 215 (with
v
0
= 0):
2
2
1
2
2
1
2
.
2
y
y
gt
t
g
y
y
gt
t
g
′
′=
′
¡
′=
=
¡
=
Plugging in
y
=
h
and
y'
= 0.50
h
, and dividing these two equations, we obtain
t
t
h
g
h
g
′
=
=
2 050
2
050
.
/
/
.
.
b
g
Letting
t'
=
t
– 1.00 (SI units understood) and crossmultiplying, we find
t
t
t
−
=
¡
=
−
100
050
100
1
050
.
.
.
.
which yields
t
= 3.41 s.
(b) Plugging this result into
y
gt
=
1
2
2
we find
h
= 57 m.
(c) In our approach, we did not use the quadratic formula, but we did “choose a root”
when we assumed (in the last calculation in part (a)) that
050
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 Spring '06
 Stoler
 Physics, Quadratic equation, one second, coordinate origin, 57 M

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