Ejercicios De Lanzamiento De Proyectiles
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Publicado: 30 de octubre de 2012
CHAPTER THREE SOLUTIONS
Chapter Three Readings
Brancazio, P, "The Trajectory of a Fly Ball," The Physics Teacher, January 1985,
p. 20.
Brown, R.A., "Maximizing the Range of a Projectile", The Physics Teacher, 30, 344, 1992
Drake, S. and MacLachlan, J., "Galileo's Discovery of the Parabolic Trajectory," Scientific American, March 1975, p. 102.
Review
[pic]
(a) Thelength of the height of the triangle is given by
h = 260.8 sin 50° = 199.8 ft.
The side DB has length
DB = 260.8 cos 50° = 167.6 ft,
leaving the side
AD = 311.4 - 167.6 = 143.8 ft.
The angle q = tan-1 = tan-1 = 35.7° and
AC = [(AD)2 + h2]1/2 = [(143.8)2 + (199.8)2] 1/2 = 246.2 ft.
The remaining side of the triangle (from C to A) isthus 246.2 ft in a direction 35.7° west of south.
(b) The area of the triangle is given by
Area = bh =
= 3.111 x 104 ft2 () = 0.714 acres.
[pic]
3.1 Your sketch should be drawn to scale, and should look somewhat like that pictured at the right. The angle from the westward direction, q, can be measured to be 4.34° north of west, and the distance from your sketch canbe converted according to the scale used to be 7.92 m.
[pic]
3.2 Your sketch when drawn to scale should look somewhat like the one at the right. The distance R and the angle q can be measured to give, upon use of your scale factor, the values of R = 421 ft at about 3° below the horizontal.
3.3 (a) Drawing these to scale and maintaining their respective directions yields aresultant of 5.2 m at an angle of 60° above the x axis.
(b) Maintain the direction of A, but reverse the direction of B by 180°. The resultant is 3.0 m at an angle of 30° below the x axis.
(c) Maintain the direction of B, but reverse the direction of A. The resultant is 3.0 m at an angle of 150° with respect to the + x axis.
(d) Maintain the direction of A, reversethe direction of B, and multiply its magnitude by two. The resultant is 5.2 m at an angle of 60° below the + x axis.
3.4 (a) Carefully draw, to scale, a vector 3.00 units long along the x direction, and from the tip of this vector, draw another of length 4.00 units in the negative y direction. The resultant is the length, to scale, of the vector drawn from the tail of the first to the tip ofthe second. This will be a vector 5.00 units long and at an angle of 53° below the x axis.
(b) In this case, the second vector, - B, will be in the + y direction. The resultant will still be 5 units long, but at an angle of 53° above the +x direction.
[pic]
3.5 The displacement vectors A = 8.00 m westward and B = 13.0 m north can be drawn to scale as at the right. The vectorC represents the displacement that the man in the maze must undergo to return to his starting point. The scale used to draw the sketch can be used to find C to be 15.3 m and the angle q can be measured to be about 58°.
[pic]
3.6 The vector diagram sketched for this problem should look like the one shown at the right. The initial displacement A = 100 m and the resultant R= 175 m are both known. In order to reach the end point of the run following the initial displacement, the jogger must follow the path shown as B. The distance can be found from the scale used for your sketch and the angle q measured. The results should be about 83 m at 33° north of west.
3.7 A total distance of = 8.07 m
with an angle of tan-1= 42.0° south of east.
3.8 Theperson would have to walk
3.10 sin(25.0°) = 1.31 km north, and 3.10 cos(25.0°) = 2.81 km east.
3.9 (a) Her resultant x (east-west) displacement is -3.00 + 0 + 6.00 = 3.00 blocks, while her resultant y (north-south) displacement is 0 + 4.00 + 0 = 4.00 blocks. Her resultant displacement is then 5.00 blocks at 53.1° north of east.
(b) Her total distance is 3.00 + 4.00 +...
Chapter Three Readings
Brancazio, P, "The Trajectory of a Fly Ball," The Physics Teacher, January 1985,
p. 20.
Brown, R.A., "Maximizing the Range of a Projectile", The Physics Teacher, 30, 344, 1992
Drake, S. and MacLachlan, J., "Galileo's Discovery of the Parabolic Trajectory," Scientific American, March 1975, p. 102.
Review
[pic]
(a) Thelength of the height of the triangle is given by
h = 260.8 sin 50° = 199.8 ft.
The side DB has length
DB = 260.8 cos 50° = 167.6 ft,
leaving the side
AD = 311.4 - 167.6 = 143.8 ft.
The angle q = tan-1 = tan-1 = 35.7° and
AC = [(AD)2 + h2]1/2 = [(143.8)2 + (199.8)2] 1/2 = 246.2 ft.
The remaining side of the triangle (from C to A) isthus 246.2 ft in a direction 35.7° west of south.
(b) The area of the triangle is given by
Area = bh =
= 3.111 x 104 ft2 () = 0.714 acres.
[pic]
3.1 Your sketch should be drawn to scale, and should look somewhat like that pictured at the right. The angle from the westward direction, q, can be measured to be 4.34° north of west, and the distance from your sketch canbe converted according to the scale used to be 7.92 m.
[pic]
3.2 Your sketch when drawn to scale should look somewhat like the one at the right. The distance R and the angle q can be measured to give, upon use of your scale factor, the values of R = 421 ft at about 3° below the horizontal.
3.3 (a) Drawing these to scale and maintaining their respective directions yields aresultant of 5.2 m at an angle of 60° above the x axis.
(b) Maintain the direction of A, but reverse the direction of B by 180°. The resultant is 3.0 m at an angle of 30° below the x axis.
(c) Maintain the direction of B, but reverse the direction of A. The resultant is 3.0 m at an angle of 150° with respect to the + x axis.
(d) Maintain the direction of A, reversethe direction of B, and multiply its magnitude by two. The resultant is 5.2 m at an angle of 60° below the + x axis.
3.4 (a) Carefully draw, to scale, a vector 3.00 units long along the x direction, and from the tip of this vector, draw another of length 4.00 units in the negative y direction. The resultant is the length, to scale, of the vector drawn from the tail of the first to the tip ofthe second. This will be a vector 5.00 units long and at an angle of 53° below the x axis.
(b) In this case, the second vector, - B, will be in the + y direction. The resultant will still be 5 units long, but at an angle of 53° above the +x direction.
[pic]
3.5 The displacement vectors A = 8.00 m westward and B = 13.0 m north can be drawn to scale as at the right. The vectorC represents the displacement that the man in the maze must undergo to return to his starting point. The scale used to draw the sketch can be used to find C to be 15.3 m and the angle q can be measured to be about 58°.
[pic]
3.6 The vector diagram sketched for this problem should look like the one shown at the right. The initial displacement A = 100 m and the resultant R= 175 m are both known. In order to reach the end point of the run following the initial displacement, the jogger must follow the path shown as B. The distance can be found from the scale used for your sketch and the angle q measured. The results should be about 83 m at 33° north of west.
3.7 A total distance of = 8.07 m
with an angle of tan-1= 42.0° south of east.
3.8 Theperson would have to walk
3.10 sin(25.0°) = 1.31 km north, and 3.10 cos(25.0°) = 2.81 km east.
3.9 (a) Her resultant x (east-west) displacement is -3.00 + 0 + 6.00 = 3.00 blocks, while her resultant y (north-south) displacement is 0 + 4.00 + 0 = 4.00 blocks. Her resultant displacement is then 5.00 blocks at 53.1° north of east.
(b) Her total distance is 3.00 + 4.00 +...
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