fisica1-youn-e-freedman-exercicios-resolvidos

3.90: As in the previous problem, the horizontal distance x in terms of the angles is
⎛ gx ⎞ 1
tanθ
= tan( θ + φ)
−
⎜
.
2 2
2
⎟
⎝ v0 ⎠ cos ( θ + φ)
2
Denote the dimensionless quantity gx / 2v
by β ; in this case
2
(9.80 m/s )(60.0 m)cos30.0°
β =
= 0.2486.
2
2(32.0 m/s)
The above relation can then be written, on multiplying both sides by the product
cosθ cos( θ + φ),
β cosθ
sinθ
cos( θ + φ)
= sin( θ + φ)cosθ
− ,
cos( θ + φ)
and so
β cosθ
sin( θ + φ)cosθ
− cos( θ + φ)sinθ
= .
cos( θ + φ)
The term on the left is sin(( θ + φ)
−θ
) = sinφ,
so the result of this combination is
0
sinφ cos( θ + φ)
= β cosθ.
Although this can be done numerically (by iteration, trial-and-error, or other methods),
the expansion sin a cosb
= 1 (sin( a + b)
+ sin( a − b )) allows the angle φ to be isolated;
2
specifically, then
1
(sin(2φ + θ ) + sin( −θ
)) = β cosθ
,
2
with the net result that
sin( 2φ + θ ) = 2β
cosθ
+ sinθ.
a) For θ = 30° , and β as found above, φ = 19. 3°
and the angle above the horizontal is
θ + φ = 49 .3°.
For level ground, using β = 0.2871,
gives φ = 17 .5°.
b) For θ = −30°
, the
same β as with θ = 30°
may be used (cos 30° = cos( −30°
)),
giving φ = 13. 0°
and
φ +θ = −17 .0°.