fisica1-youn-e-freedman-exercicios-resolvidos

3.60: a) This may be done by a direct application of the result of Problem 3.59; with
α
0
= −40°
, substitution into the expression for x gives 6.93 m.
b)
c) Using ( 14. 0 m − 19 . m) instead of h in the above calculation gives x = 6.3 m , so the
man will not be hit.
3.61: a) The expression for the range, as derived in Example 3.10, involves the sine of
twice the launch angle, and
sin (2(90° −α 0
)) = sin (180° − 2α
0
) = sin 180°
cos 2α
0
− cos180°
sin 2α0
= sin 2α
0
,
and so the range is the same. As an alternative, using sin( 90° −α0 ) = cosα
and
cos( 90° −α
0)
= sinα0
in the expression for the range that involves the product of the sine
and cosine of α
0
gives the same result.
2
v sin 2α
0
b) The range equation is R =
g
. In this case, v = 2.2 0
m/s and R = 0.25 m .
2
2
Hence sin 2α = (9.8 m/s )(0.25 m)/(2.2 m/s ), or sin 2α = 0. 5062 ; and α = 15. 2°
or
74 .8° .