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Composition Setup
 Deck of Cards
idbigdeck
 Wiki Markup
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
 Toggle Cloak
idsysa
 System: 
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idsysa
The ball will be treated as a .
 Toggle Cloak
idinta
 Interactions: 
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idinta
External influence from the earth (gravity).
 Toggle Cloak
idmoda
 Models: 
 Cloak
idmoda
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.
 Toggle Cloak
idappa
 Approach: 
 Cloak
idappa
 Toggle Cloak
iddiaga
  Diagrammatic Representation 
 Cloak
iddiaga
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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diaga
diaga
 Toggle Cloak
idmatha
  Mathematical Representation 
 Cloak
idmatha
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.
 Cloak
matha
matha
 Cloak
appa
appa
 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.
 Toggle Cloak
idappb
 Approach: 
 Cloak
idappb
 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.
 Toggle Cloak
idappc
 Approach: 
 Cloak
idappc