fisica1-youn-e-freedman-exercicios-resolvidos

3.89: a) The trajectory of the projectile is given by Eq. (3.27), with α 0
= θ + φ, and the
equation describing the incline is y = x tanθ.
Setting these equal and factoring out the
x = 0 root (where the projectile is on the incline) gives a value for x
0;
the range
measured along the incline is
2
2
⎡2v
⎤
⎡
0
cos ( θ + φ)
⎤
x / cosθ
= ⎢ ⎥[tan(
θ + φ)
− tan θ]
⎢
.
cos
⎥
⎣ g ⎦
⎣ θ ⎦
b) Of the many ways to approach this problem, a convenient way is to use the same
sort of "trick", involving double angles, as was used to derive the expression for the range
along a horizontal incline. Specifically, write the above in terms of α = θ + φ, as
2
⎡ 2v
⎤
2
R = ⎢
0 [sinα
cosα
cosθ
− cos α sin ].
2 ⎥
cos θ
θ
⎣ g ⎦
The dependence on α and hence φ is in the second term. Using the identities
2
sinα cosα
= (1/ 2)sin 2α
and cos α = (1/ 2)(1 + cos2α
), this term becomes
( 1/ 2)[cosθ
sin 2α
− sin θ cos2α − sin θ]
= (1/ 2)[sin(2α − θ)
− sin θ].
This will be a maximum when sin( 2α − θ)
is a maximum, at 2 α − θ = 2φ
+ θ = 90°
, or
φ = 45°
− θ / 2. Note that this reduces to the expected forms when θ = 0 (a flat incline,
φ = 45° and when θ = −90°
(a vertical cliff), when a horizontal launch gives the greatest
distance).