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h2. Part A Suppose you are throwing a baseball. You release the ball with a perfectly horizontal velocity of 5.0 m/s at a height of 1.5 m above the ground. How far will the ball travel horizontally from the instant it leaves your hand until the instant it first contacts the ground? System: The ball will be treated as a [point particle] subject to an influence from the earth (gravity). Models: The ball is in projectile motion, so we model the x-component of the ball's motion as [One-Dimensional Motion with Constant Velocity|1-D Motion (Constant Velocity)] and the y-component as [One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)]. Approach: The first thing to do is to sketch the situation, which allows us to summarize the givens and unknowns and also to set up a coordinate system. !baseball2.png! In the problem statement, we are told that _h_ = 1.5 m (as drawn in the picture) and we are asked for _d_. By drawing coordinate axes into our picture we have denoted the positive _x_ and _y_ directions. We have not yet chosen the origin, however (the axes can be placed wherever you wish on the picture to avoid clutter). We will take that step now. We choose our origin such that the position _x_ = 0 m is the location at which the ball leaves the hand. The location _y_ = 0 m is the level of the ground. With these choices made, we can summarize the givens (along with our traditional choice that _t_~i~ = 0 s): {panel:title=givens}{latex}
Card
labelPart A

Part A

Suppose you are throwing a baseball. You release the ball with a perfectly horizontal velocity of 5.0 m/s at a height of 1.5 m above the ground.

Excerpt

How far will the ball travel horizontally from the instant it leaves your hand until the instant it first contacts the ground?

Solution

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System:
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The ball will be treated as a .

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Interactions:
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External influence from the earth (gravity).

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Models:
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The ball is in projectile motion, so we model the x-component of the ball's motion as One-Dimensional Motion with Constant Velocity and the y-component as One-Dimensional Motion with Constant Acceleration.

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Approach:

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Diagrammatic Representation

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The first thing to do is to sketch the situation, which allows us to summarize the givens and unknowns and also to set up a coordinate system.

Image Added

In the problem statement, we are told that h = 1.5 m (as drawn in the picture) and we are asked for d. By drawing coordinate axes into our picture we have denoted the positive x and y directions. We have not yet chosen the origin, however (the axes can be placed wherever you wish on the picture to avoid clutter). We will take that step now. We choose our origin such that the position x = 0 m is the location at which the ball leaves the hand. The location y = 0 m is the level of the ground.

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Mathematical Representation

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With the choice of coordinates made, we can summarize the givens (along with our traditional choice that ti = 0 s):

Panel
titlegivens
Latex
\begin{large}\[ t_{\rm i} = \mbox{0 s} \] \[ x_{\rm i} = \mbox{0 m} \]\[ x = d \] \[ y_{\rm i} = \mbox{1.5 m} \] \[ y = \mbox{0 m}\]\[v_{x} = \mbox{5.0 m/s} \]\[v_{y,{\rm i}} = \mbox{0 m/s} \] \[ a_{y} = -\mbox{9.8 m/s}^{2}\]\end{large}
{latex}{panel} {note}It is important to note that the phrase *perfectly horizontal velocity of 5.0 m/s* implies that the full velocity
Note

It is important to note that the phrase perfectly horizontal velocity of 5.0 m/s implies that the full velocity (5.0

m/s)

is

directed

along

the

x-axis,

with

zero

y-component

for

the

initial

velocity.

This

phrasing

is

extremely

common

in

physics.

You

will

also

encounter

the

perpendicular

case

of

a

"perfectly

vertical

velocity".

It

is

also

worth

remarking

that

although

5.0

m/s

is

the

velocity

at

the

instant

of

release

(clearly

the

ball's

initial

velocity

for

the

freefall

trajectory

of

interest)

we

have

written

_

v

_~_

x

_~

=

5.0

m/s

rather

than

_

v

_~_

x

_

,

i~

i =

5.0

m/s.

This

is

not

a

typo,

because

the

_

x

_

direction

is

subject

to

the

1-D

Motion

with

Constant

Velocity

model

(recall

_

a

_~_

x

_~

=

0).

Because

the

_

x

_

velocity

is

constant,

it

does

not

require

labels

for

initial

or

final

states.

{note}

We

are

asked

for

_

d

_

,

which

appears

in

the

_

x

_

direction

givens.

For

this

reason,

we

should

first

consider

the

Laws

of

Change

available

for

the

_

x

_

direction.

Because

of

the

simplicity

of

the

1-D

Motion

with

Constant

Velocity

model,

there

is

only

one

available

Law

of

Change:

{

Latex
}
\begin{large} \[ x = x_{\rm i} + v_{x} (t-t_{\rm i}) \] \end{large}
{latex}

We

cannot

solve

this

equation,

however,

because

we

do

not

know

_

x

_

or

_

t

_

.

We

therefore

proceed

to

use

an

extremely

useful

technique.

We

will

use

the

_

y

_

direction

to

solve

for

the

time

_

t

_

.

Once

we

have

this,

we

can

use

it

in

the

_

x

_

direction

equation

to

obtain

the

information

requested

by

the

problem.

{

Note
}

In

projectile

motion,

if

the

direction

which

explicitly

contains

the

desired

unknown

quantity

does

not

yield

solvable

equations

using

the

problem

givens,

it

is

*

extremely

*

likely

that

you

should

solve

the

other

direction

for

time.

{note}

In

the

_

y

_

direction

we

have

four

Laws

of

Change

to

consider.

The

most

direct

way

to

proceed

is

to

use:

{

Latex
}
\begin{large} \[ y = y_{\rm i} + v_{y,{\rm i}}(t-t_{\rm i}) + \frac{1}{2} a_{y}(t-t_{\rm i})^{2}\]\end{large}
{latex}

which

at

first

looks

messy,

but

after

substituting

zeros

becomes:

{

Latex
}
\begin{large} \[ 0 = y_{\rm i} + \frac{1}{2} a_{y}t^{2} \] \end{large}
{latex}

which

is

solved

to

give:

{

Latex
}
\begin{large} \[ t = \pm \sqrt{\frac{-2y_{\rm i}}{a_{y}}} \] \end{large}
{latex}

and

we

must

choose

the

plus

sign

since

we

have

already

set

up

the

problem

with

the

ball

released

at

_

t

_

=

0

s.

This

time

can

be

substituted

directly

into

the

_

x

_

direction

Law

of

Change

to

give:

{

Latex
}
\begin{large} \[ x = d = v_{x}\sqrt{\frac{-2y_{\rm i}}{a_{y}}} \] \end{large}
{latex}

To

be

clear,

we

show

the

substitution:

{

Latex
}
\begin{large} \[ d = (\mbox{5.0 m/s})\sqrt{\frac{-2(\mbox{1.5 m})}{-\mbox{9.8 m/s}^{2}}} = \mbox{2.8 m} \] \end{large}
{latex} {note}Note that the negative sign under the square root was canceled by the negative _y_ acceleration. When you see a negative sign appear under a square root, you should always check that it is canceled by the algebraic signs of the given quantities. If it does not cancel, it is a sign of a math error! Such warnings are extremely valuable when checking work.{note} Part B Suppose you are throwing a baseball. 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. How fast was the ball moving when you released it? System & Models: As in Part A. Approach: This question works basically the same as Part A, except that the givens are slightly different. In this part, we have (assuming the same coordinate system as was used in Part A): {panel:title=givens}{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} {note}Note that although we are not given the _x_ component of the initial velocity, the phrase "perfectly horizontal velocity" still tells us that the _y_ component is zero initially.{note} Once again, we have an unknown that requires the _x_ direction's Law of Change, but we do not have enough information. We proceed in exactly the same fashion as in Part A to find the same answer for the time: {latex}\begin{large} \[ t = \sqrt{\frac{-2y_{\rm i}}{a_{y}}} \] \end{large}{latex} We then rearrange the _x_ direction equation and substitute for the time: {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} Part C
Note

Note that the negative sign under the square root was canceled by the negative y acceleration. When you see a negative sign appear under a square root, you should always check that it is canceled by the algebraic signs of the given quantities. If it does not cancel, it is an indication of a math error! Such warnings are extremely valuable when checking work.

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Card
labelPart B

Part B

Suppose you are throwing a baseball. 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. How fast was the ball moving when you released it?

Solution

System, Interactions and Models: As in Part A.

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Approach:

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Card
labelPart C

Part C

Suppose a certain major-league pitcher releases a fastball with a perfectly horizontal velocity of 95 mph. The ball is released at a height of 6.0 feet above the ground and travels 60.0 feet before being caught by the catcher. At what height above the ground should the cathcer place his glove?

System, Interactions and Models: As in Parts A and B.

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Approach:

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