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| } This problem assumes familiarity with the earlier example problem, [ ]. {note} |
Again
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suppose
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you
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are
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throwing
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a
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baseball,
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just
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as
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we
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assumed
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in
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Parts
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A
...
and
...
B
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of
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the
...
...
...
.
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You
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release
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the
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ball
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with
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a
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perfectly
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horizontal
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velocity
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at
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a
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height
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of
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1.5
...
m
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above
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the
...
ground.
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The
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ball
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travels
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5.0
...
m
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horizontally
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from
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the
...
instant
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it
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leaves
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your
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hand
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until
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the
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instant
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it
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first
...
contacts
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the
...
ground.
...
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How |
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fast |
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is the |
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ball |
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moving |
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just |
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before |
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it |
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impacts |
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the |
...
ground? |
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Solution
System, Interactions and Models:
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As
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in
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Parts
...
A
...
and
...
B
...
of
...
the
...
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The givens are the same as in Part B of the earlier example. Defining the same coordinate system as was used in that Part, we have:
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example|Throwing a Baseball 1 (The Basics)]
Approach: The givens are the same as in Part B of the earlier example. Defining the same coordinate system as was used in that Part, we have:
{panel:title=givens}{latex}
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Further,
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we
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can
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follow
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the
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same
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steps
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as
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we
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did
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in
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Part
...
B
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from
...
before
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to
...
find
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the
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x-velocity.
| Latex |
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}\begin{large} \[ v_{x} = \frac{x}{t} = x \sqrt{\frac{a_{y}}{-2y_{\rm i}}} = (\mbox{5.0 m})\sqrt{\frac{-\mbox{9.8 m/s}^{2}}{-2(\mbox{1.5 m})}} = \mbox{9.0 m/s} \] \end{large}{latex} |
This
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time,
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however,
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we
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are
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not
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done
...
yet.
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The
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x-velocity
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is
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only
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half
...
of
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the
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story.
...
When
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the
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ball
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is
...
first
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released,
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its
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velocity
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is
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entirely
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in
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the
...
x
...
direction,
...
but
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during
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its
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flight
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it
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acquires
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a
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y-component.
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When asked how fast something is traveling, we generally make no distinction between x- and y-components.
...
Instead,
...
we
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want
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to
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know
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the
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size
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of
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the
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total
...
velocity.
...
To
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find
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the
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total
...
velocity,
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we
...
must
...
use
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the
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Pythagorean
...
theorem:
| Latex |
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PICTURE {latex}\begin{large} \[ v = \sqrt{v_{x}^{2} + v_{y}^{2}} \] \end{large}{latex} |
There
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are
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many
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ways
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to
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find
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the
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y-component
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of
...
the
...
velocity.
...
One
...
possibility
...
is
...
to
...
use
...
the
...
equation:
| Latex |
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} \begin{large} \[ v_{y}^{2} = v_{y,{\rm i}}^{2} + 2 a_{y}(y-y_{\rm i}) \] \end{large}{latex} |
which,
...
after
...
substituting
...
zeros
...
gives:
| Latex |
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} \begin{large} \[ v_{y} = \pm \sqrt{-2 a_{y}y_{\rm i}} = \pm \sqrt{-2(-\mbox{9.8 m/s}^{2})(1.5 m)}= \pm \mbox{5.4 m/s} \] \end{large}{latex} |
In
...
this
...
case,
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we
...
must
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choose
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the
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negative
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sign.
...
This
...
is
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true
...
because
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the
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ball
...
is
...
clearly
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moving
...
downward
...
(as
...
opposed
...
to
...
upward)
...
just
...
before
...
striking
...
the
...
ground.
...
Thus,
...
the
...
vertical
...
component
...
of
...
the
...
ball
...
is
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negative
...
in
...
our
...
coordinate
...
system.
| Latex |
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{latex}\begin{large} \[ v_{y} = - \mbox{5.4 m/s}\]\end{large}{latex} |
We
...
therefore
...
find
...
that:
| Latex |
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} \begin{large} \[ v = \sqrt{(\mbox{9.0 m/s})^{2} + (-\mbox{5.4 m/s})^{2}} = \mbox{10.5 m/s} \] \end{large}{latex} |
Note
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that
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we
...
did
...
NOT
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choose
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between
...
plus
...
or
...
minus
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this
...
time.
...
The
...
Pythagorean
...
theorem
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does
...
not
...
have
...
one.
...
The
...
reason
...
for
...
this
...
is
...
that
...
the
...
Pythagorean
...
theorem
...
gives
...
lengths
...
(when
...
dealing
...
with
...
its
...
orginal
...
application),
...
or
...
in
...
the
...
case
...
of
...
vectors
...
like
...
velocity
...
or
...
acceleration,
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magnitudes
...
.
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Magnitudes
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can
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never
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be
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negative.
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This
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leads
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us
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to
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an
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important
...
aside.
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Whenever
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the
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Pythagorean
...
theorem
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is
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being
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used,
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we
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are
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clearly
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working
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with
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vectors
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in
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two
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or
...
more
...
dimensions.
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Once
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we
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have
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moved
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beyond
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one
...
dimension,
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it
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is
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impossible
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to
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unambiguously
...
report
...
the
...
direction
...
of
...
a
...
vector
...
by
...
assigning
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a
...
plus
...
or
...
minus
...
sign
...
to
...
the
...
magnitude.
...
The
...
baseball
...
of
...
this
...
problem
...
is
...
a
...
good
...
example.
...
It
...
is
...
clearly
...
moving
...
to
...
the
...
right
...
(positive
...
in
...
our
...
coordinate
...
system)
...
and
...
down
...
(negative
...
in
...
our
...
coordinate
...
system).
To illustrate the appropriate way to report direction, let us now find the velocity of the ball just before contacting the ground.
We have the vector triangle:
| Note |
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Note that we have removed the negative sign from the y-component. This is correct, because the arrow is used in the drawing to denote the direction. An arrow and a negative sign can result in confusion, in just the same way as reporting a velocity of " – 2 m/s west" can result in confusion. |
We can now find the angle shown in the figure by using trigonometry. One possibility is to use:
| Latex |
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Thus it is both positive _and_ negative, in a sense. To illustrate the appropriate way to report direction, let us now find the *velocity* of the ball just before contacting the ground. We have the vector triangle: PICTURE {note}Note that we have removed the negative sign from the y-component. This is correct, because the *arrow* is used in the drawing to denote the direction. An arrow *and* a negative sign can result in confusion, in just the same way as reporting a velocity of " -- 2 m/s west" can result in confusion.{note} We can now find the angle shown in the figure by using trigonometry. One possibility is to use: {latex}\begin{large} \[ \tan^{-1}\left(\frac{\mbox{5.4 m/s}}{\mbox{9.0 m/s}}\right) = 31^{\circ} \] \end{large}{latex} {info}Can you get the same result for θ using the arccosine or arcsine functions?{info} Reporting the vector is a little tricky here, since we are not told which direction the ball is thrown (north? south? east? west?). The best we can do is to say the final velocity is 10.5 m/s at 31° *below the horizontal*. |
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Can you get the same result for θ using the arccosine or arcsine functions? |
Reporting the vector is a little tricky here, since we are not told which direction the ball is thrown (north? south? east? west?). The best we can do is to say the final velocity is 10.5 m/s at 31° below the horizontal.
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