Game Development Reference
In-Depth Information
Notice that the result is negative, indicating an upwards velocity. If
we give the ball bearing this initial velocity, we might wonder how long
it takes for the bearing to come back down to its initial position. Using
Equation (11.16) and letting ∆x = 0, we have
v
0
+ 2a∆x
a
−v
0
±
t =
(−79.2 ft/s)
2
+ 2(32 ft/s
2
)(0 ft)
32 ft/s
2
−(−79.2 ft/s) ±
=
=
79.2 ft/s ±
(−79.2 ft/s)
2
32 ft/s
2
=
79.2 ft/s ± 79.2 ft/s
32 ft/s
2
= 0 or 4.95 s.
It's no surprise that t = 0 is a solution; we were solving for the time values
when the ball bearing was at its initial position.
Examine the graph in Figure 11.12, which plots the position and veloc-
ity of an object moving under constant velocity a with an initial velocity
v
0
, where v
0
and a have opposite signs. Let's make three key observations.
Although we use terms such as “height,” which are specific to projectile
motion, similar statements are true anytime the signs of v
0
and a are op-
posite.
Figure 11.12
Projectile motion
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