3 MOTION IN 3 DIMENSIONS 3.11 Projectile motion(^0 0)
s = (x, y, z) from its launch point satisfies [see Eq. (3.35)]
s = v 0
t +
1
a t^2. (3.43)
2
Making use of Eqs. (3.41) and (3.42), the x-, y-, and z-components of the above
equation are written
x = v 0 cos θ t, (3.44)
y = 0, (3.45)
z = v 0 sin^ θ^ t^ −^
1
g t^2 , (3.46)
2
respectively. Note that the projectile moves with constant velocity, vx = dx/dt =
v 0 cos θ, in the x-direction (i.e., horizontally). This is hardly surprising, since
there is zero component of the projectile’s acceleration along the x-axis. Note,
further, that since there is zero component of the projectile’s acceleration along
the y-axis, and the projectile’s initial velocity also has zero component along this
axis, the projectile never moves in the y-direction. In other words, the projectile’s
trajectory is 2 - dimensional, lying entirely within the x-z plane. Note, finally, that
the projectile’s vertical motion is entirely decoupled from its horizontal motion.
In other words, the projectile’s vertical motion is identical to that of a second pro-
jectile launched vertically upwards, at t = 0, with the initial velocity v 0 sin θ (i.e.,
the initial vertical velocity component of the first projectile)—both projectiles will
reach the same maximum altitude at the same time, and will subsequently strike
the ground simultaneously.
Equations (3.44) and (3.46) can be rearranged to give
1 g x^2 2
(^) z = x tan θ −
2 2
sec
0
θ. (3.47)
As was first pointed out by Galileo, and is illustrated in Fig. 17 , this is the equa-
tion of a parabola. The horizontal range R of the projectile corresponds to its
x-coordinate when it strikes the ground (i.e., when z = 0). It follows from the
above expression (neglecting the trivial result x = 0 ) that
2 v 2
R =
g
sin θ cos θ =
v 2
g
sin 2θ. (3.48)
v