fisica1-youn-e-freedman-exercicios-resolvidos

2 2
v sin α
=
2g
0 0 0
3.71: a) The height above the player’s hand will be = 0.40 m , so the
maximum height above the floor is 2.23 m. b) Use of the result of Problem 3.59 gives
3.84 m. c) The algebra is the same as that for Problems 3.58 and 3.62. The distance y is
3 .05 m − 1.83 m = 1.22 m , and
v
2cos
(9.80 m/s
2
v y
2g
)(4.21m)
2
2
0
=
=
2
35°
((4.21m) tan 35° −1.22 m)
8.65 m/s.
d) As in part (a), but with the larger speed,
2 2
2
1.83 m + (8.65 m/s) sin 35°
/ 2(9.80 m/s ) = 3.09 m.
The distance from the basket is the distance from the foul line to the basket, minus half
the range, or
2
2
4.21m − (8.655 m/s) sin 70°
/ 2(9.80 m/s ) = 0.62 m.
Note that an extra figure in the intermediate calculation was kept to avoid roundoff error.
3.72: The initial y-component of the velocity is v0 y
= 2gy
, and the time the pebble is in
2
flight is t = 2y
/ g . The initial x-component is v0 x
= x / t = x g / 2y
. The magnitude of
the initial velocity is then
0 y
and the angle is arctan( ) arctan (2y
/ x)
v
v0
x
v
0
=
= .
2
x g
⎛ x ⎞
2gy
+ = 2gy
1+
⎜ ⎟
2y
⎝ 2y
⎠
2
,