fisica1-youn-e-freedman-exercicios-resolvidos

2.80: If the speed of the flowerpot at the top of the window is v 0 , height h of the
window is
h = v
ave
t
= v t
0
2
+ ( 1 2) gt , or v0
=
h
t
− (1 2) gt.
The distance l from the roof to the top of the window is then
2
2
0 =
2
v ((1.90 m) /(0.420 s) − (1 2)(9.80 m s )(0.420 s))
l = =
2
2 g
2(9.80 m s )
0.310 m.
An alternative but more complicated algebraic method is to note that t is the difference
between the times taken to fall the heights l + h and h, so that
2(
l + h)
2l
2
t = − , gt 2 + l = l + h.
g g
Squaring the second expression allows cancelation of the l terms,
which is solved for
2
2
(1 2) gt + 2 gt l 2 = h,
1 ⎛ h ⎞
l = ⎜ − (1 2) gt ⎟ ,
2g
⎝ t ⎠
which is the same as the previous expression.
2
( 5.00 m s)
2.81: a) The football will go an additional
v 2
=
2
( )
= 1.27 m above the window, so
2 g 2 9.80 m s
the greatest height is 13.27 m or 13.3 m to the given precision.
2
b) The time needed to reach this height is 2(13.3m) (9.80 m s ) = 1.65s.
2