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{
Note
}

This

problem

assumes

familiarity

with

the

earlier

example

problem,

[

Throwing

a

Baseball

1

(The

Basics)

]

.

{note}

Again

...

suppose

...

you

...

are

...

throwing

...

a

...

baseball,

...

just

...

as

...

we

...

assumed

...

in

...

Parts

...

A

...

and

...

B

...

of

...

the

...

earlier

...

example

...

.

...

You

...

release

...

the

...

ball

...

with

...

a

...

perfectly

...

horizontal

...

velocity

...

at

...

a

...

height

...

of

...

1.5

...

m

...

above

...

the

...

ground.

...

The

...

ball

...

travels

...

5.0

...

m

...

horizontally

...

from

...

the

...

instant

...

it

...

leaves

...

your

...

hand

...

until

...

the

...

instant

...

it

...

first

...

contacts

...

the

...

ground.

...

Excerpt

How

...

fast

...

is the

...

ball

...

moving

...

just

...

before

...

it

...

impacts

...

the

...

ground?

...

Solution

System, Interactions and Models:

...

As

...

in

...

Parts

...

A

...

and

...

B

...

of

...

the

...

earlier example

Toggle Cloak
idapp
Approach:

Cloak
idapp

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
titlegivens
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}
Latex
\begin{large}\[ t_{\rm i} = \mbox{0 s} \] \[ x_{\rm i} = \mbox{0 m} \]\[ x = d = \mbox{5.0 m}\] \[ y_{\rm i} = \mbox{1.5 m} \] \[ y = \mbox{0 m}\]\[v_{y,{\rm i}} = \mbox{0 m/s} \] \[ a_{y} = -\mbox{9.8 m/s}^{2}\]\end{large}
{latex}{panel}

Further,

...

we

...

can

...

follow

...

the

...

same

...

steps

...

as

...

we

...

did

...

in

...

Part

...

B

...

from

...

before

...

to

...

find

...

the

...

x-velocity.

{
Latex
}\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

...

time,

...

however,

...

we

...

are

...

not

...

done

...

yet.

...

The

...

x-velocity

...

is

...

only

...

half

...

of

...

the

...

story.

...

When

...

the

...

ball

...

is

...

first

...

released,

...

its

...


velocity

...

is

...

entirely

...

in

...

the

...

x

...

direction,

...

but

...

during

...

its

...

flight

...

it

...

acquires

...

a

...

y-component.

...

Image Added

When asked how fast something is traveling, we generally make no distinction between x- and y-components.

...

Instead,

...

we

...

want

...

to

...

know

...

the

...

size

...

of

...

the

...

total

...

velocity.

...

To

...

find

...

the

...

total

...

velocity,

...

we

...

must

...

use

...

the

...

Pythagorean

...

theorem:

Image Added

Latex


PICTURE

{latex}\begin{large} \[ v = \sqrt{v_{x}^{2} + v_{y}^{2}} \] \end{large}{latex}

There

...

are

...

many

...

ways

...

to

...

find

...

the

...

y-component

...

of

...

the

...

velocity.

...

One

...

possibility

...

is

...

to

...

use

...

the

...

equation:

{
Latex
} \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
} \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,

...

we

...

must

...

choose

...

the

...

negative

...

sign.

...

This

...

is

...

true

...

because

...

the

...

ball

...

is

...

clearly

...

moving

...

downward

...

(as

...

opposed

...

to

...

upward)

...

just

...

before

...

striking

...

the

...

ground.

...

Thus,

...

the

...

vertical

...

component

...

of

...

the

...

ball

...

is

...

negative

...

in

...

our

...

coordinate

...

system.

Latex
  

{latex}\begin{large} \[ v_{y} = - \mbox{5.4 m/s}\]\end{large}{latex}

We

...

therefore

...

find

...

that:

{
Latex
} \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

...

that

...

we

...

did

...

NOT

...

choose

...

between

...

plus

...

or

...

minus

...

this

...

time.

...

The

...

Pythagorean

...

theorem

...

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,

...

magnitudes

...

.

...

Magnitudes

...

can

...

never

...

be

...

negative.

...

This

...

leads

...

us

...

to

...

an

...

important

...

aside.

...

Whenever

...

the

...

Pythagorean

...

theorem

...

is

...

being

...

used,

...

we

...

are

...

clearly

...

working

...

with

...

vectors

...

in

...

two

...

or

...

more

...

dimensions.

...

Once

...

we

...

have

...

moved

...

beyond

...

one

...

dimension,

...

it

...

is

...

impossible

...

to

...

unambiguously

...

report

...

the

...

direction

...

of

...

a

...

vector

...

by

...

assigning

...

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:

Image Added

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.

We can now find the angle shown in the figure by using trigonometry. One possibility is to use:

Latex
  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*.









Info

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.

Cloak
app
app