Game Development Reference
In-Depth Information
86 592 (ft/s)
2
32 ft/s
2
= ±
≈ ±
294.3 ft/s
32 ft/s
2
≈ ±9.197 s.
The square root in Equation (11.16) introduces the possibility for two so-
lutions. We always use the root that results in a positive value for t.
30
Naturally, a person in the business of dropping ball bearings from great
heights is interested in how much damage he can do, so the next logical ques-
tion is, “How fast is the ball bearing traveling when it hits the sidewalk?”
To answer this question, we plug the total travel time into Equation (11.13):
v(t) = v
0
+ at = 0 ft/s + (32 ft/s
2
)(9.197 s) = 294.3 ft/s.
If we ignore wind resistance, at the moment of impact, the ball bearing is
traveling at a speed that covers a distance of roughly a football field in one
second! You can see why the things we are doing in our imagination are
illegal in real life. Let's keep doing them.
Now let's assume that instead of just dropping the ball bearing, we give
it an initial velocity (we toss it up or down). It was our free choice to decide
whether up or down is positive in these examples, and we have chosen +x
to be the downward direction, so that means the initial velocity will be
negative. What must the initial velocity be in order for the ball bearing to
stay in the air only a few seconds longer, say a total of 12 seconds? Once
again, we'll first manipulate Equation (11.15) to get a general solution; this
time we'll be solving for v
0
:
∆x = v
0
t + (1/2)at
2
,
−v
0
t = −∆x + (1/2)at
2
,
v
0
= ∆x/t − (1/2)at.
Solving for initial
velocity
And now plugging in the numbers for our specific problem, we have
v
0
= ∆x/t − (1/2)at
= (1 353 ft)/(12.0 s) − (1/2)(32 ft/s
2
)(12.0 s)
= 112.8 ft/s − 192 ft/s
= −79.2 ft/s.
30
The negative root tells us the other point where the infinite parabola containing the
ball bearing's trajectory crosses the sidewalk.
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