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Part A

Suppose the roommate drops the keys straight down. The keys are released at rest from a height h= 5.0 m above the outstretched hand of the forgetful student. How long are the keys in the air from the time the roommate releases them until the instant they land in the student's hand?

Solution

Although the keys do receive an acceleration from the roommate while they are in contact with his hand, the instant they leave the roommate's hand gravity takes over. The observational evidence of this is that the keys will immediately begin to slow down, indicating the action of a downward force.

}{composition-setup}


{excerpt:hidden=true}Keys moving in 1D freefall with or without initial velocity.{excerpt}

A college student with a third floor dorm room is exiting the dorm when he suddenly realizes he has forgotten his keys. Rather than run back to the room, he calls his roommate and goes to stand under the dorm room window.  The roommate brings the keys to the window.  

{deck:id=partdeck}
{card:label=Part A}

h3. Part A

  Suppose the roommate drops the keys straight down.  The keys are released at rest from a height _h_= 5.0 m above the outstretched hand of the forgetful student.  How long are the keys in the air from the time the roommate releases them until the instant they land in the student's hand?

h4. Solution

{toggle-cloak:id=sysa} *System:*  {cloak:id=sysa}The keys will be treated as a point particle.{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}The earth has an influence on the keys, giving them a constant downward acceleration of magnitude _g_.{cloak}

{toggle-cloak:id=moda} *Model:*  {cloak:id=moda}[One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)].  The keys are in freefall with a constant acceleration from gravity.  We will choose our axis such that the positive direction is upward.{cloak}

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

{cloak:id=appa}
We are looking for a law of change that will make use of _h_ and _g_ and _v_~i~ to find _t_.  
{note}The word "rest" is a keyword in physics problems.  The phrase "released at rest" in the second sentence of Part 1 indicates that _v_~i~=0, which is a necessary piece of information if the problem is to be solved.{note}
The appropriate equation is:

{latex}\begin{large} $x = x_{\rm i} + v_{\rm i} t + \frac{1}{2} a t^{2}$\end{large}{latex}

The appropriate substitutions are:

{latex}\begin{large}\(x - x_{\rm i} = - h \)\end{large}{latex}
{note}The negative is important here, implying (for our choice of axis) that the keys moved downward.{note}
{latex}\begin{large}\(a = -g \)\end{large}{latex}
{warning}In this text, as in the vast majority of physics resources, _g_ is a magnitude (_g_ = + 9.8 m/s ^2^).{warning}
which, when combined with _v_~i~ = 0 gives:

{latex}\begin{large}\( - h = - \frac{1}{2} g t^{2} \) \end{large}{latex}

or:

{latex}\begin{large}\[ t = \pm \sqrt{\frac{2h}{g}}\] \end{large}{latex}
The plus sign is chosen, since the keys will be caught after they are dropped, so:

{latex}\begin{large}\[ t = \sqrt{\frac{2h}{g}} = 1.0 \;{\rm s}\] \end{large}{latex}
{note}The negative root is unphysical in _this_ problem, since the keys were not in freefall until released at _t_=0.  The negative root _could_ have physical meaning in another problem, however, if the system was in freefall both before and after _t_ = 0.{note}

{cloak}
{card}
{card:label=Part B}

h3. Part B

 Suppose that, instead of dropping the keys, the roommate tosses them straight up with an initial speed of 3.3 m/s.  The keys are released from a height _h_= 5.0 m above the outstretched hand of the forgetful student.  How long are the keys in the air from the time the roommate releases them until the instant they land in the student's hand?

h4. Solution

*System, Interactions and Model:*  The system, interactions and model are the same as for Part A.

{note}Although the keys do receive an acceleration from the roommate while they are in contact with his hand, the _instant_ they leave the roommate's hand gravity takes over.  The observational evidence of this is that the keys will immediately begin to slow down, indicating the action of a downward force.{note}

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

{cloak:id=appb}

We will illustrate two separate but completely equivalent ways to do this problem.  The first way is faster, but requires familiarity with the quadratic equation.  The second way avoids the quadratic equation by making use of symmetry, but it requires more physical insight.

{deck:id=quaddeck}
{card:label=With Quadratic}

h5. With Quadratic

The problem is identical to Part 1 except that _v_~i~ is not zero.  Thus, the equation that we have to solve is:

{latex}\begin{large}\( - h = v_{\rm i}t -\frac{1}{2}gt^{2}\)\end{large}{latex}

 {color}

{latex}\begin{large}\[ t = \frac{v_{\rm i} + \sqrt{v_{\rm i}^{2} + 2gh}}{g}= 1.4\;{\rm s}\]\end{large}{latex}

{note}The negative root 

The negative root is unphysical in this problem, since the keys were not in freefall until released at t=0. The negative root could have physical meaning in another problem, however, if the system was in freefall both before and after t = 0.

is unphysical in _this_ problem, since the keys were not in freefall until released at _t_=0.  The negative root _could_ have physical meaning in another problem, however, if the system was in freefall both before and after _t_ = 0.{note}

{card:With Quadratic}
{card:label=Without Quadratic}

h5.  Without Quadratic

|!keysprojectile.gif!|

Freefall (and later, projectile) problems can often be usefully broken into two parts and analyzed in a mathematically straightforward fashion.  The point at which we will separate the problem into two parts is the point of maximum height. As we will see, this is a useful choice because the velocity goes to zero at that point.  We begin by analyzing the upward motion of the keys.
  
Considering only the trip up to maximum height (shown in blue in the figure at right), we know both the initial and final velocities.  Since we also have the acceleration (gravity), this information allows us to use the simplest available Law of Change in our model:
{latex}\begin{large}\(v = v_{\rm i} + a t \)\end{large}{latex}

{latex}.

We are now ready to solve for the time to fall down from the peak (the part of the path shown in red in the figure at right).  The solution proceeds exactly as in Part 1, so we use that result.  The only change is that the keys are now falling from a total height of:

{latex}\begin{large}\[ h_{\rm peak} = h + \frac{v_{\rm i}^{2}}{2g} \] \end{large}{latex}

The same answer obtained using the quadratic formula.

{latex}
{tip}The same answer obtained using the quadratic formula.{tip}

{card:Without Quadratic}
{deck:quaddeck}
{cloak:appb}
{card:Part B}
{card:label=Part C}


h3. Part C

As a final illustration, suppose that the roommate throws the keys _downward_ with an initial speed of 3.3 m/s.  The keys are released from a height _h_= 5.0 m above the outstretched hand of the forgetful student.  How long are the keys in the air from the time the roommate releases them until the instant they land in the student's hand?

h4. Solution

*System, Interactions and Model:* As in the previous parts.

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

{cloak:id=appc}

As in Part B, we will illustrate two separate but completely equivalent ways to do this problem.  The first way is one step, but requires familiarity with the quadratic equation.  The second way avoids the quadratic equation, but it requires two steps and stronger physical insight.

{deck:id=quaddeck2}
{card:label=With Quadratic}

h5. With Quadratic

The solution is exactly the same as for Part B.  The only difference is that _v_~i~ = -3.3 m/s instead of +3.3 m/s.  Thus, we find:

{latex}\begin{large}\[ t = \frac{v_{\rm i} + \sqrt{v_{\rm i}^{2} + 2gh}}{g}= 0.73\;{\rm s}\]\end{large}{latex}

{card:With Quadratic}
{card:label=Without Quadratic}

h5. Without Quadratic

This time, the motion has no peak point at which the keys are turning around.  Thus, we cannot break up the motion as we did in Part B.  Instead, we use a two step process.  It is very valuable in freefall (and projectile) problems to know the initial and final velocities.  In this case, we only have the initial velocity as a given, but we can find the final velocity as a first step toward solving the problem.  To do this, we must recognize that without time the only Law of Change for our model that can be solved for the final velocity is:

{latex}

\begin{large}\[ v^{2} = v_{\rm i}^{2} + 2a(x-x_{\rm i})\]\end{large}{latex}

The plus or minus sign here is very, very important. When taking a square root to find velocity, you must remember to use your knowledge of the problem to decide which sign is the correct one!

 caught.

{latex}\begin{large}\[ v = - \sqrt{v_{\rm i}^{2} + 2gh} \]\end{large}{latex}

It is worth noting that although we could not use the method of breaking the motion at the peak that we used in Part B here in Part C, the method of finding the final velocity that we have just used here in Part C would have worked very well in Part B! Perhaps you find this method simplest, but do not forget that it is up to you to remember to assign the appropriate plus or minus to the final velocity. It is very easy to forget this step.