Page History

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.  

h4. Part 1

  Suppose the roommate drops the keys straight down from rest.  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?

System:  {color:maroon}The keys will be treated as a point particle.  The earth has an influence on the keys, giving them a constant downward acceleration of magnitude _g_.{color}

Model:  {color:maroon}One-Dimensional Motion with 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.{color}

Approach: {color:maroon} 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 "from rest" in the first 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:{color}
{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}
{color:maroon}which, when combined with _v_~i~ = 0 gives:{color}

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

{color:maroon}or:{color}
{latex}\begin{large}\[ t = \sqrt{\frac{2h}{g}} = 1.0 \;{\rm s}\] \end{large}{latex}

h4. Part 2

 Suppose 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?

System \& Model:  The system and model are the same as for Part 1.

{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}

Approach:  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.

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}
This is a quadratic equation.  It is a good idea to rearrange it:
{latex}\begin{large}\( \frac{1}{2}gt^{2} - v_{\rm i}t - h = 0\)\end{large}{latex}
so that it is clear that we have an equation of the form
{latex}\begin{large}\(At^{2}+Bt+C=0\)\end{large}{latex}
if we choose:
{latex}\begin{large}\(A = 1/2 g\)\end{large}{latex}
{latex}\begin{large}\(B = -v_{\rm i}\)\end{large}{latex}
{latex}\begin{large}\(C = - h\)\end{large}{latex}
Using these assignments in the quadratic equation gives:
{latex}\begin{large}\[ t = \frac{v_{\rm i} \pm \sqrt{v_{\rm i}^{2} + 2gh}}{g} = 1.4 \;{\rm s}\]\end{large}{latex}
To obtain a positive time, the positive sign must be chosen since the radical expression will clearly evaluate to be larger than _v_~i~.  
{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}

h5.  Without Quadratic

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.  This is a good choice because the velocity goes to zero at that point.  This problem will illustrate the power of this fact.  We begin by analyzing the upward motion of the keys.
  
Considering only the trip up to maximum height, we now 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}
so, using _a_=-_g_ and _v_=0:
{latex}\begin{large}\[ t_{\rm up} = \frac{v_{\rm i}-v}{g} = \frac{v_{\rm i}}{g}\]\end{large}{latex}
We have the time, which is what we wanted, but we must now make a slight detour and find how far the keys travel during their upward motion.  The reason for this is that (as you will see) we will need that height to find the time the keys take to fall down from the peak to the student's hand.  Luckily, this distance is easy to find.  There are many equations that can be solved for it, but we will choose the one that is mathematically simplest:
{latex}\begin{large}\[ x = x_{\rm i} + \frac{v+v_{\rm i}}{2} t \]\end{large}{latex}
Now if we use our expression for _t_~up~ and the fact that _v_ = 0 at the peak, we find that the distance traveled up from the roommate's hand is:
{latex}\begin{large}\[ x-x_{\rm i} = \frac{v_{i}^{2}}{2} g \]\end{large}{latex}.

We are now ready to solve for the time to fall down from the peak.  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}\[ h_{\rm peak} = h + \frac{v_{i}^{2}}{2} g \] \end{large}{latex}
Then, using the answer to Part 1, the time to fall is:
{latex}\begin{large}\[t_{\rm down}=\sqrt{\frac{2hg+v_{\rm i}^{2}}{g^{2}}}=\frac{\sqrt{v_{\rm i}^{2}+2gh}}{g}\]\end{large}{latex}

The total time is:
{latex}\begin{large}\[t=t_{\rm up}+t_{\rm down}=\frac{v_{\rm i}+\sqrt{v_{\rm i}^{2}+2gh}}{g}=1.4\;{\rm s}\] \end{large}{latex}
{tip}The same answer obtained using the quadratic formula.{tip}

h4. Part 3