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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.

{

Solution

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.

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

\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}

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:

}\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.

In

the

y

direction

we

have

four

Laws

of

Change

to

consider.

The

most

direct

way

to

proceed

is

to

use:

}\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:

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

which

is

solved

to

give:

}\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:

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

To

be

clear,

we

show

the

substitution:

}\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 an indication of a math error!  Such warnings are extremely valuable when checking work.{note}

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