fisica1-youn-e-freedman-exercicios-resolvidos

3.59: a) In Eq. (3.27), the change in height is y = −h
. This gives a quadratic equation in
x, the solution to which is
2
v0
cos α ⎡
0 2 2gh
⎤
x = ⎢tan
α0
+
2 ⎥
g ⎣ v0
cos α0
⎦
v0
cosα0
2 2
= [ v0
sin α0
+ v0
sin α0
+ 2gh
].
g
If h = 0 , the square root reduces to v sin α , and x = R
0 0
. b) The expression for x
2
2
becomes x = 10.2 m) cosα
+ [sin α + sin α 0.98]
(
0 0
0
+
The angle α = 90°
corresponds to the projectile being launched straight up, and there
0
is no horizontal motion. If α 0 , the projectile moves horizontally until it has fallen the
distance h.
0 =
c) The maximum range occurs for an angle less than 45 ° , and in this case the angle is
about 36 ° .