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.
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: 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.
Model: 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.
Approach: We are looking for a law of change that will make use of h and g and vi to find t.
The word "rest" is a keyword in physics problems. The phrase "from rest" in the first sentence of Part 1 indicates that vi=0, which is a necessary piece of information if the problem is to be solved.
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:
\begin
(x - x_
= - h )\end
The negative is important here, implying (for our choice of axis) that the keys moved downward.
a = -g
In this text, as in the vast majority of physics resources, g is a magnitude (g = + 9.8 m/s 2).
which, when combined with vi = 0 gives:
\begin
( - h = - \frac
g t^
) \end
or:
\begin
[ t = \sqrt{\frac
{g}} = 1.0 \;
] \end
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.
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.
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.
With quadratic: The problem is identical to Part 1 except that vi is not zero. Thus, the equation that we have to solve is:
\begin
( - h = v_
t -\frac
gt^
)\end
This is a quadratic equation. It is a good idea to rearrange it:
\begin
( \frac
gt^
- v_
t - h = 0)\end
so that it is clear that we have an equation of the form
\begin
(At^
+Bt+C=0)\end
if we choose:
A = 1/2 g
B = -vi
C = - h
Using these assignments in the quadratic equation gives:
\begin
[ t = \frac{v_
\pm \sqrt{v_
^
+ 2gh}}
= 1.4 \;
]\end
To obtain a positive time, the positive sign must be chosen since the radical expression will clearly evaluate to be larger than vi.
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.
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:
\begin
(v = v_
+ a t )\end
so, using a=-g and v=0:
\begin
[ t = \frac{v_
-v}
= \frac{v_
}
]\end