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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


!vectriangle.png!

{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:

!veltriangle.png!

{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