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{
Wiki Markup
Composition Setup
 Note
This problem assumes familiarity with the earlier example problem, Throwing a Baseball 1 (The Basics).
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
}{composition-setup}

{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|Throwing a Baseball 1 (The Basics)].  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?{excerpt}

h4. Solution

*System, Interactions and Models:*  As in Parts A and B of the [earlier example|Throwing a Baseball 1 (The Basics)]

{toggle-cloak:id=app} *Approach:* 

{cloak:id=app}

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 


!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*.

{cloak:app}

 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