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Physics Challenge for
Teachers and Students
Boris Korsunsky, Column Editor
Weston High School, Weston, MA 02493
[email protected]
Solution to September 2014 Challenge
w The second half of the flight
A rock is thrown vertically. During the last second of its flight, it
covers half the total distance traveled. What is the maximum possible flight time for the rock? Assume that the acceleration due to
gravity is 10 m/s2.
(Adapted from Olympiadnye Zadachi po Fizike, M. Bakunov, S. Biragov, R&C Dynamics, Moscow, 2006)
Again note that both terms on the RHS of Eq. (2) are
manifestly positive.
Despite the devious wording of the problem, quite a few
readers solved the problem correctly. We received many
solutions in which the contributors chose to explore the
problem further. Here is one such solution.
Qualitative analysis
For maximum flight time we want to arrange both
that the distance traveled in the last second is “large”
and that the average speed before that was “small.”
Together these characteristics will produce a long
flight time before the last second begins. The first
of these requirements will be best met if the rock is
thrown upward, slows to momentary rest, and then
falls some distance before the last second of the flight
begins.The second suggests that the rock should
spend most of the time before the last second falling
from its highest point so that the last second begins
with as high a speed as possible. As we’ll see, that is
precisely the case.
Solution: Let v1 be the velocity at the beginning of the
last second (the “intermediate velocity”) and assume,
as explained above, that it is downward, which we will
take to be negative, we find the distance d traveled during the last second to be
1
(1)
d = −v1t + g t 2 ,
2
where t  1 s and g  10 m/s2. Note that both terms
on the RHS of Eq. (1) are manifestly positive.
Next, assuming (as explained above) that the rock was
initially thrown upward with velocity v0 > 0, the distance traveled before the beginning of the final second
(specified in the problem statement to be equal to d) is
the sum of the distances traveled while rising and while
falling,
The Physics Teacher ◆ Vol. 52, 2014
d=
v02 v12
+ .
2g 2g (2)
Finally, the total flight time T is given by the sum of the
rising and falling times before the last second plus one
second. That is,
v
−v
T = 0 + 1 + t.
(3)
g
g
We extremize T by setting dT/dv1 = 0, which yields
dv
0
= 1.
(4)
dv1
Eliminating d between Eqs. (1) and (2) and solving for v0,
we find
(5)
v0 = g2 t 2 − 2 gt v1 − v12 ,
so that Eq. (4) becomes
v1(v1 + 2gt) = 0,
(6)
which has two roots,
v1 = 0 and v1 = –2gt.
(7)
The first of these is the upper boundary of the range for
which our analysis is valid. We reject it both because the
derivative in Eq. (4) is not well-defined at that point and
because it violates our intuition that v1 should be strictly
and substantially negative. Thus, with the help of Eq. (5),
we find
v1 = – 2gt = –20 m/s and v0 = gt = 10 m/s,
and, from Eq. (3),
Tmax = T v1 = −2 gt = 4t = 4 s.
Discussion
(8)
(9)
The total time of flight T, the “last second distance” d,
and the initial velocity v0 are all uniquely determined
by the intermediate velocity v1. The reverse, however, is not true. For instance, there is a range of initial
velocities, specifically from v0 = 200 / 3 m/s ≈ 8.16 m/s to
v0 = 200 m/s ≈ 14.14 m/s (the derivation of these values,
and the rest of the analysis here is left as an exercise for
the reader), for which the conditions of the problem
are satisfied for three different values of v1 and associated values of d and T.
In case (a), the rock is shot upward, and the total
distance D it travels is represented by the area A +
(B + C) under the graph, given by
The figure below illustrates the multi-valued nature of
the dependence of the total time of flight on the initial
velocity. Note especially that, while the maximum time
of flight occurs for an initial velocity of 10 m/s with
where k is a proportionality constant, common to
all triangles in both figures, by the similarity of the
formers. Moreover, according to the problem, the
distance C traveled during the last second is equal
to B + A, and by the similarity of triangles, we have
1) an intermediate velocity of –20 m/s and a total time
of flight of 4 s (5 m up and 20 m down followed by
25 m down in the final second),
Time of Flight versus Initial Velocity
3.500
3.000
In case (b), on the other hand, the rock is shot
downward, and from Fig. (b) it is readily seen that
2.500
2.000
D = k[(t –T)2 – T2] and C = k[(t – 1 – T)2 – T2],
1.500
which, together with D = 2C, yields
1.000
0.500
t = T + 2 + T 2 + 2, t = T + 2 − T 2 + 2, T ≤ 0.
0.000
-10.000
0.000
Equating D to 2C and solving for t, we find
Since we are looking for the maximum t, the second root is of no interest, while the first one attains
a maximum t = 4 s at T = 1 s.
4.000
-20.000
C = k [T2 + (t –1 – T)2].
t = T + 2 + −T 2 + 2, t = T + 2 − −T 2 + 2, T ≥ 0.
4.500
-30.000
D = k [T2 + (t – T)2],
10.000
20.000
30.000
Initial Velocity (m/s)
there are two other, lesser total time-of-flight scenarios
that also satisfy the problem statement. One has
2) an intermediate velocity of 0 and a total time of
flight of 2 s (5 m up followed by 5 m down)
and the other has
3) an intermediate velocity of 20/3 < 6.67 m/s and a
total time of flight of 4/3 < 1.33 s (25/9 m up followed by another 20/9 m up and 5/9 m down during
the final second).
(Contributed by John Mallinckrodt, Cal Poly Pomona,
Pomona, CA)
We again may ignore the second root. As to the
first, it is a strictly increasing function of T which
attains a maximum of 2 + 2 at T = 0 ; this seems
intuitively clear, for the closer T to zero, the lower
the initial speed and the longer it takes the rock
to cover the first half of the total distance covered.
Combining both results, we conclude that the maximum flight time is four seconds. It is interesting to
note that the answer is independent of the acceleration due to gravity.
(Submitted by Joseph Rizcallah, Sagesse High School,
Ain Saadeh, Lebanon)
We also recognize the following contributors:
Salvatore Basile (Università degli Studi di Palermo,
Palermo, Italy)
Philip Blanco (Grossmont College, El Cajon, CA)
Speed
Speed
Don Easton (Lacombe, Alberta, Canada)
Gerald E. Hite (TAMUG, Galveston, TX)
Art Hovey (Galvanized Jazz Band, Milford, CT)
C
C
A
B
Time
Time David Jones (Florida International University,
0
t -1 t
0
t
T
T
t -1
Miami, FL)
(a)
(b)
José Ignacio Íñiguez de la Torre (Universidad de
The figure above shows the speed-time graph of the
Salamanca, Salamanca, Spain)
motion of the rock, with T being the time (possibly
Jarrett L. Lancaster (University of North Carolina at
negative if the rock is shot downward, as shown in Fig.
Greensboro, Greensboro, NC)
(b)) the rock reaches the maximum height and t is the
Matthew W. Milligan (Farragut High School, Knoxtotal time of flight.
ville, TN)
And here is an alternative solution that uses a geometric approach:
The Physics Teacher
◆ Vol. 52, 2014
Daniel Mixson (Naval Academy Preparatory School, Newport, RI)
Carl E. Mungan (U. S. Naval Academy, Annapolis,
MD)
Pascal Renault (John Tyler Community College, Midlothian,
VA)
Ryan Yancey (Lancaster High School, Lancaster, CA)
Guidelines for contributors:
– We ask that all solutions, preferably in Word format, be
submitted to the dedicated email address challenges@
aapt.org. Each message will receive an automatic acknowledgment.
– The subject line of each message should be the same as
the name of the solution file.
– The deadline for submitting the solutions is the last day of
the corresponding month.
464
The Physics Teacher ◆ Vol. 52, 2014
– Each month, a representative selection of the successful solvers’ names will be published in print and
on the web. – If your name is—for instance—Bilbo Baggins,
please name the file “Nov14Baggins” (do not include your first initial) when submitting the November solution.
– If you have a message for the Column Editor, you
may contact him at [email protected];
however, please do not send your solutions to this
address.
As always, we look forward to your contributions
and hope that they will include not only solutions
but also your own Challenges that you wish to submit for the column. Boris Korsunsky, Column Editor