98 Chapter 2 Force and Motion
Next, let’s find the location of the ball. Let’s split it into constituent parts in the x and y
directions:
x = v
1
x
t
y = v
1
y
t − ½ gt
2
Now, if there were only some way to eliminate the time variable in this second equa-
tion. Perhaps we should rearrange that first equation!
t =
x
—
v
1
x
Substituting that into the second equation, we get the following:
y = v
1
y
x
(
—
)
v
1
x
− ½ g
x
(
—
)
v
1
x
2
This is actually a quadratic function, and it will show a parabola when plotted. The ori-
gin is at the point where the ball leaves the hand.
From this equation, you can tell where the thrown ball will land. Actually, we can take
the term
x
(
—
)
v
1
x
out of this equation like so:
y =
x
—
v
1
x
(
v
1
y
− ½ g
x
(
—
)
v
1
x
)
And given that we know that the ball’s landing ...