Sep 2008
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Physics Challenge for Teachers and Students Solution to September 2008 Challenge w Half and Half Challenge: A rock is launched vertically. During the last second of the flight, the rock covers one-half of the entire distance covered during the flight. What is the maximum possible duration of the flight? (Adapted from M. Bakunow and S. Biragov, Problems from Physics Olympiads [R&C Dynamics, Moscow, 2006], in Russian.)
Solution: Assuming that air drag is negligible, let us first consider what happens if the rock lands at the same altitude as it is launched from. In that case, half of the total distance is covered on the way up and half on the way down. Then, by symmetry, the rock takes 1 s to go up and 1 s to go down, for a total flight time of 2 s. The question thus becomes: If the rock lands at a different altitude than the one from which it is launched, is it possible to achieve a greater flight time? Well, if the rock lands at a higher altitude than its launch position (for instance, it lands on the roof of a building), then its average speed near the end of its flight is smaller than that near the beginning of its flight. In that case, it will require less than 1 s in the first part of its flight to travel the same distance that it travels in the final second of its flight. Then the total flight time is less than 2 s, which is not what we want. In contrast, similar arguments make it clear that the flight time will be longer than 2 s if the landing position is below the launch height (say we throw the rock off the edge of a cliff ). Having determined that, the next question is: Should the rock be thrown upward, downward, or be simply dropped over the edge of the cliff? As I
have already argued, we want the average speed in the first part of the trip to be as small as possible. That way the rock will take as long as possible to cover the same distance that it covers in the last second of its flight. It is therefore clear that we should throw the rock directly upward, so that it will slow down, instantaneously stop, and then only gradually regain speed. To summarize, we now know that the trajectory of the rock must look something like what is sketched below. 1 2
3
I have divided the motion of the rock into three phases. During interval 1, the rock rises vertically a distance of y1 in a time of t1; in interval 2, the rock falls vertically a distance of y2 in a time of t2; and finally in interval 3, the rock falls vertically a further distance of y3 L in a time of t3 = 1 s. The statement of the problem requires that y1 + y2 = L. Our goal is to maximize the total flight time T t1 + t2 + t3. With the problem now clearly set up, we proceed to solve it in an organized fashion. Elementary ki-
The Physics Teacher ◆ Vol. 46, 2008
nematics tells us that y1 = 12 gt12 and
y2 = 12 gt22
so that 1 2
(
(1)
(2)
)
g t12 + t22 = L.
(3) Kinematics also tells us that the speed of the rock as it passes the dashed line segment in the figure is u = gt2. In that case, during the third segment the distance traveled is
L = ut3 + 12 gt32 = gt 2 + 12 g ,
(4)
where in the second step I substituted t3 = 1 s and suppressed the units of s and s2 (at the slight cost of dimensional consistency). Equating Eqs. (3) and (4), and multiplying through by 2/g, we obtain
t12 + t 22 = 2t 2 + 1 ⇒ t1 = 2t 2 + 1 − t 22 (5) and thus the total flight time is
T = t1 + t 2 + t3 = 2t 2 + 1 − t 22 + t 2 + 1.
(6)
(As a check, note that t2 = 0 ➯ t1 = 1 s and T = 2 s, which is the flat-ground result.) To maximize T, we differentiate it with respect to t2 and set the result to zero. After a bit of algebra, one finds t2 = 2 s . Equation (5) then implies that t1 = 1 s, and consequently the total flight time is 4 s, which is the solution of the problem. The interested reader can go on to show that the
The Physics Teacher ◆ Vol. 46, 2008
cliff is 39.2 m high, the ball is launched upward off it with a speed of 9.8 m/s, it rises upward to a height of y1 = 4.9 m, the dashed line is located 14.7 m below the top of the cliff, the ball crosses that point with a speed of u = 19.6 m/s, and the ball hits the ground with a speed of 29.4 m/s. (Contributed by Carl E. Mungan, U.S. Naval Academy, Annapolis, MD) We would also like to recognize the following contributors: Phil Cahill (Lockheed Martin Corporation, Rosemont, PA) Fredrick P. Gram (Cuyahoga Community College, Cleveland, OH) Fernando Ferreira (Universidade da Beira Interior, Covilhã, Portugal) Art Hovey (Milford, CT) Peter Jung, student (Science Academy of South Texas, Mission, TX) Stephen McAndrew (Trinity Grammar School, Summer Hill, NSW, Australia) Matthew W. Milligan (Farragut High School, Knoxville, TN) Dawit Zewdie, student (Albert Einstein High School, Silver Spring, MD) Many thanks to all contributors and we hope to hear from you in the future! Please send correspondence to:
Boris Korsunsky [email protected]
Solution: Assuming that air drag is negligible, let us first consider what happens if the rock lands at the same altitude as it is launched from. In that case, half of the total distance is covered on the way up and half on the way down. Then, by symmetry, the rock takes 1 s to go up and 1 s to go down, for a total flight time of 2 s. The question thus becomes: If the rock lands at a different altitude than the one from which it is launched, is it possible to achieve a greater flight time? Well, if the rock lands at a higher altitude than its launch position (for instance, it lands on the roof of a building), then its average speed near the end of its flight is smaller than that near the beginning of its flight. In that case, it will require less than 1 s in the first part of its flight to travel the same distance that it travels in the final second of its flight. Then the total flight time is less than 2 s, which is not what we want. In contrast, similar arguments make it clear that the flight time will be longer than 2 s if the landing position is below the launch height (say we throw the rock off the edge of a cliff ). Having determined that, the next question is: Should the rock be thrown upward, downward, or be simply dropped over the edge of the cliff? As I
have already argued, we want the average speed in the first part of the trip to be as small as possible. That way the rock will take as long as possible to cover the same distance that it covers in the last second of its flight. It is therefore clear that we should throw the rock directly upward, so that it will slow down, instantaneously stop, and then only gradually regain speed. To summarize, we now know that the trajectory of the rock must look something like what is sketched below. 1 2
3
I have divided the motion of the rock into three phases. During interval 1, the rock rises vertically a distance of y1 in a time of t1; in interval 2, the rock falls vertically a distance of y2 in a time of t2; and finally in interval 3, the rock falls vertically a further distance of y3 L in a time of t3 = 1 s. The statement of the problem requires that y1 + y2 = L. Our goal is to maximize the total flight time T t1 + t2 + t3. With the problem now clearly set up, we proceed to solve it in an organized fashion. Elementary ki-
The Physics Teacher ◆ Vol. 46, 2008
nematics tells us that y1 = 12 gt12 and
y2 = 12 gt22
so that 1 2
(
(1)
(2)
)
g t12 + t22 = L.
(3) Kinematics also tells us that the speed of the rock as it passes the dashed line segment in the figure is u = gt2. In that case, during the third segment the distance traveled is
L = ut3 + 12 gt32 = gt 2 + 12 g ,
(4)
where in the second step I substituted t3 = 1 s and suppressed the units of s and s2 (at the slight cost of dimensional consistency). Equating Eqs. (3) and (4), and multiplying through by 2/g, we obtain
t12 + t 22 = 2t 2 + 1 ⇒ t1 = 2t 2 + 1 − t 22 (5) and thus the total flight time is
T = t1 + t 2 + t3 = 2t 2 + 1 − t 22 + t 2 + 1.
(6)
(As a check, note that t2 = 0 ➯ t1 = 1 s and T = 2 s, which is the flat-ground result.) To maximize T, we differentiate it with respect to t2 and set the result to zero. After a bit of algebra, one finds t2 = 2 s . Equation (5) then implies that t1 = 1 s, and consequently the total flight time is 4 s, which is the solution of the problem. The interested reader can go on to show that the
The Physics Teacher ◆ Vol. 46, 2008
cliff is 39.2 m high, the ball is launched upward off it with a speed of 9.8 m/s, it rises upward to a height of y1 = 4.9 m, the dashed line is located 14.7 m below the top of the cliff, the ball crosses that point with a speed of u = 19.6 m/s, and the ball hits the ground with a speed of 29.4 m/s. (Contributed by Carl E. Mungan, U.S. Naval Academy, Annapolis, MD) We would also like to recognize the following contributors: Phil Cahill (Lockheed Martin Corporation, Rosemont, PA) Fredrick P. Gram (Cuyahoga Community College, Cleveland, OH) Fernando Ferreira (Universidade da Beira Interior, Covilhã, Portugal) Art Hovey (Milford, CT) Peter Jung, student (Science Academy of South Texas, Mission, TX) Stephen McAndrew (Trinity Grammar School, Summer Hill, NSW, Australia) Matthew W. Milligan (Farragut High School, Knoxville, TN) Dawit Zewdie, student (Albert Einstein High School, Silver Spring, MD) Many thanks to all contributors and we hope to hear from you in the future! Please send correspondence to:
Boris Korsunsky [email protected]
