119
Example: a car going over a cliff
Question: The police find a car at a distance w=20 m from the
base of a cliff of height h=100 m. How fast was the car going
when it went over the edge. Solve the problem symbolically first,
then plug in the numbers.
Solution: Let’s choose y pointing up and x pointing away from
the cliff. The car’s vertical motion was independent of its horizon-
tal motion, so we know it had a constant vertical acceleration of
a=-g=-9.8 m/s
2
. The time it spent in the air is therefore related to
the vertical distance it fell by the constant-acceleration equation
.
y=
1
2
a
y
.
t
2
,
or
–h=
1
2
(–g)
.
t
2
.
Solving for
.
t gives
.
t=2h
g
.
Since the vertical force had no effect on the car’s horizontal
motion, it had a
x
=0, i.e. constant horizontal velocity. We can
apply the constant-velocity equation
v
x
=
.
x
.
t
,
i.e.
v
x
=w
.
t
.
We now substitute for
.
t to find
v
x
=w/2h
g
,
which simplifies to
v
x
=wg
2h
.
Plugging in numbers, we find that the car’s speed when it went
over the edge was 4 m/s, or about 10 mi/hr.
Projectiles move along parabolas
What type of mathematical curve does a projectile follow through
space. To find out, we must relate x to y, eliminating t. The reasoning is very
similar to that used in the example above. Arbitrarily choosing x=y=t=0 to
be at the top of the arc, we conveniently have x=
.
x, y=
.
y, and t=
.
t, so
y=–
1
2
a
y
t
2
x
=v
x
t
We solve the second equation for t=x/v
x
and eliminate t in the first equa-
tion:
y=–
1
2
a
y
x
v
x
2
.
Since everything in this equation is a constant except for x and y, we
conclude that y is proportional to the square of x. As you may or may not
recall from a math class, y
.
x
2
describes a parabola.
Each water droplet follows a pa-
rabola. The faster drops’ parabolas
are bigger.
w
h
vx=.
x
y
Section 6.2Coordinates and Components
A parabola can be defined as the
shape made by cutting a cone paral-
lel to its side. A parabola is also the
graph of an equation of the form
y
.
x
2
.
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