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

The appropriate substitutions are:
xf - xi = - h

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:

Unknown macro: {latex}

\begin

Unknown macro: {large}

( - h = - \frac

Unknown macro: {1}
Unknown macro: {2}

g t^

) \end

or:

Unknown macro: {latex}

\begin

Unknown macro: {large}

( t = \sqrt{\frac

Unknown macro: {2h}

{g}} )= 1.0 s \end

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