
62 Classical Mechanics
© 2010 Taylor & Francis Group, LLC
yx
g
v
x=−tan
cos
α
α2
0
22
2
and
y = x tan θ
where θ is the slope of the inclined plane with respect to the horizontal. Eliminating y, we
obtain
Rx
vg
P
=
=−
()sec
sincos ta
ec
θ
αα
0
2
0
22
22
The maximum range along the incline, for a given θ, occurs at
(α)
max
= θ/2 + π/4 (3.14)
and the maximum range itself is
Rvg
max
(sin )sec=−
−
0
21 2
1
.
Figure 3.3 illustrates the case where a projectile is launched to a higher level; θ and h are
positive. For the case where a projectile is launched to a lower level, θ and h are negative.
Alternatively, we can find the maximum range of a projectile along a ...