v
=
2
a
Δ
y
,
were
a
= (7/3)
g
and
Δ
y
= 0.600 m. This causes the ball to reach a maximum height
h
max
(measured above the water surface) given by
h
max
=
v
2
/2
g
(see Eq. 216 again).
Thus,
h
max
= (7/3)
Δ
y
= 1.40 m.
44. Due to the buoyant force, the ball accelerates upward (while in the water) at rate
a
given by Newton’s second law:
ρ
water
Vg
–
ball
Vg
=
ball
Va
¡
ball
=
water
(1 + “
a
”)
where – for simplicity – we are using in that last expression an acceleration “
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration, Force

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