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 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
System:
Interactions:
Model:
Approach:
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?
Solution
System, Interactions and Model: The system, interactions and model are the same as for Part A.
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:
\begin
[A = \frac
g] [B = -v_
] [C = - h]\end
Using these assignments in the quadratic equation gives:
\begin
[ t = \frac{v_
\pm \sqrt{v_
^
+ 2gh}}
]\end
To obtain a positive time, the positive sign must be chosen since the radical expression will clearly evaluate to be larger than vi, so:
\begin
[ t = \frac{v_
+ \sqrt{v_
^
+ 2gh}}
= 1.4\;
]\end
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. 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:
\begin
(v = v_
+ a t )\end
so, using a=-g and v=0:
\begin
[ t_
= \frac{v_
-v}
= \frac{v_
}
]\end
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:
\begin
[ x = x_
+ \frac{v+v_{\rm i}}
t ]\end
Now if we use our expression for tup and the fact that v = 0 at the peak, we find that the distance traveled up from the roommate's hand is:
\begin
[ x-x_
= \frac{v_
^{2}}
]\end
.
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:
\begin
[ h_
= h + \frac{v_
^{2}}
] \end
Then, using the answer to Part A, the time to fall is:
\begin
[t_
=\sqrt{\frac{2hg+v_
{2}}{g
}}=\frac{\sqrt{v_
^
+2gh}}
]\end
The total time is:
\begin
[t=t_
+t_
=\frac{v_
+\sqrt{v_
^
+2gh}}
=1.4\;
] \end
The same answer obtained using the quadratic formula.
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?
Solution
System, Interactions and Model: As in the previous parts.
Approach:
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.
With Quadratic
The solution is exactly the same as for Part B. The only difference is that vi = -3.3 m/s instead of +3.3 m/s. Thus, we find:
\begin
[ t = \frac{v_
+ \sqrt{v_
^
+ 2gh}}
= 0.73\;
]\end
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:
\begin
[ v^
= v_
^
+ 2a(x-x_
)]\end
Using the familiar substitutions, this becomes:
\begin
[ v = \pm \sqrt{v_
^
+ 2gh} ]\end
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!
In this case, the minus sign must be chosen, since the keys are clearly moving downward just before they are caught.
\begin
[ v = - \sqrt{v_
^
+ 2gh} ]\end
Now that we have both the initial and final velocity, we can use the very simple Law of Change:
\begin
[ v = v_
+ at ]\end
to find:
\begin
[ t = \frac{v_
- \left(- \sqrt{v_
^
+ 2gh}\right)}
]\end
which gives:
\begin
[ t = \frac{v_
+ \sqrt{v_
^
+ 2gh}}
= 0.73\;
]\end
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.
Part D – Challenge
Can you show that the answer to Part C can be obtained from the answer to Part B by subtracting the time the keys spent above 5.0 m?