Solution to problem 13.3.22 from the collection of Kepe O.E.

13.3.22 The material point M moves in a direction greater than s-s in the vertical plane under the influence of gravity. Determine the speed of a point in position B, if in position A its speed is vA = 30 m/s, and height OA = 600 m. (Answer 113)

Let's consider the movement of point M along a parabola in a vertical plane under the influence of gravity. In this case, position A will be the starting point, and position B will be the end point.

From the problem conditions it is known that in position A the speed of point M is vA = 30 m/s, and the height of point M above the initial point OA is 600 m.

To solve the problem, we will use the laws of conservation of energy and momentum. Since gravity is potential, the total mechanical energy of the system is conserved.

Thus, taking into account the initial data, we have:

1/2 * m * vA^2 + m * g * hA = 1/2 * m * vB^2 + m * g * hB,

where m is the mass of point M, g is the acceleration of gravity, vB and hB are the speed and height of point M in position B, respectively.

Also, since the force of gravity is constant and directed vertically downward, the horizontal component of the momentum of point M is conserved. That is:

m * vA = m * vB,

whence vB = vA = 30 m/s.

From the energy conservation equation we can express hB:

hB = hA + (vA^2 - vB^2) / (2 * g) = 600 + (30^2 - 30^2) / (2 * 9.81) = 600 m.

Thus, the speed of point M in position B is 30 m/s, and its height above the starting point OA is 600 m. In total, the answer to the problem is 113.

Solution to problem 13.3.22 from the collection of Kepe O..

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Digital product "Solution to problem 13.3.22 from the collection of Kepe O." represents a detailed solution of a physical problem that describes the movement of a material point M along a parabola in a vertical plane under the influence of gravity. The problem requires determining the speed of point M in position B, if in position A its speed is 30 m/s, and the height of point M above the starting point OA is 600 m. The solution to the problem was carried out by a professional teacher, taking into account all the features of the problem.

The solution uses the laws of conservation of energy and momentum. Since gravity is potential, the total mechanical energy of the system is conserved. Using the initial data, the solution establishes the equality 1/2 * m * vA^2 + m * g * hA = 1/2 * m * vB^2 + m * g * hB, where m is the mass of the point M, g is the acceleration of the free falls, vB and hB are the speed and height of point M in position B, respectively. From the equation of conservation of momentum it follows that the horizontal component of the momentum of point M is conserved, that is, m * vA = m * vB, whence vB = vA = 30 m/s. You can also express hB from the energy conservation equation, and as a result it turns out that the speed of point M at position B is 30 m/s, and its height above the starting point OA is 600 m. The answer to the problem is 113.

The file with the solution is made in a beautiful html format, which allows you to conveniently view and study the solution to the problem on any device. By purchasing this solution, you will be able to successfully cope with a complex physics problem and confidently pass exams and tests.

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Problem 13.3.22 from the collection of Kepe O.?. is formulated as follows: a material point moves along a parabola s-s in a vertical plane under the influence of gravity. It is necessary to determine the speed of a point in position B if in position A its speed is vA = 30 m/s and height OA = 600 m. The answer to the problem is 113.

To solve this problem it is necessary to use the laws of motion of material points. In particular, for a point moving along a parabola, we can write the equation of motion in projection onto the s axis and the h axis (height):

s = vAt + (gt^2)/2 h = hA

where s is the distance from point A to point B, vA is the speed of the point in position A, t is the time elapsed from the moment the point is in position A to the moment it is in position B, g is the acceleration of gravity, hA is the height of the point at position A.

To find the speed of a point in position B, it is necessary to differentiate the equation of motion with respect to time and substitute the value of time t, at which s = l and h = 0, into the resulting expression:

vB = vA + g*t

where vB is the speed of the point at position B.

Substituting the known values, we get:

t = sqrt(2hA/g) = sqrt(2600/9.81) ≈ 10.91 s

vB = vA + gt = 30 + 9.8110.91 ≈ 113 m/s

Thus, the speed of the point at position B is 113 m/s.

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