The task of D1-20 is to determine the speed of the parachutist at the moment of landing. A parachutist of mass m begins a vertical descent from a height h = 200 m without an initial speed. Air resistance is proportional to the square of the speed and is expressed by the formula R = 3mv^2.
To solve the problem it is necessary to use the laws of mechanics. Since the skydiver is moving in a vertical direction, we can use the equation of motion of a body freely falling under the influence of gravity and air resistance:
mg - R = ma,
where m is the mass of the parachutist, g is the acceleration of gravity, R is the force of air resistance, a is the acceleration of descent.
Considering that at the moment of landing the speed of the parachutist is zero and the height h = 0, we can find the speed of the parachutist at the moment of landing by solving the equation of motion:
mg - 3mv^2 = ma,
where a = g for vertical descent.
Solving the equation, we get:
v = sqrt(g*m/3)*sqrt(2h/g),
where sqrt is the square root.
Thus, the speed of the parachutist at the moment of landing is v = sqrt(2gh/3), where g = 9.8 m/s^2 is the acceleration of gravity.
This digital product is the solution to problem D1 option 20 (D1-20), compiled by the author Dievsky V.A.
The solution to the problem describes the vertical descent of a parachutist of mass m from a height h = 200 m without an initial speed in the presence of an air resistance force proportional to the square of the speed, R = 3mv^2.
To solve the problem, the laws of mechanics were used and the answer was obtained in the form of the speed of the parachutist at the moment of landing, which is v = sqrt(2gh/3), where g = 9.8 m/s^2 is the acceleration of free fall.
By purchasing this digital product, you receive a ready-made solution to problem D1-20 from an experienced author and can use it for your educational purposes.
This digital product is a solved problem D1-20, compiled by the author Dievsky V.A. The problem is to determine the speed of a parachutist at the moment of landing during a vertical descent from a height of h = 200 m without an initial speed, in the presence of an air resistance force proportional to the square of the speed, R = 3mv^2. To solve the problem, the laws of mechanics were used, and the answer was obtained in the form of the speed of the parachutist at the moment of landing, which is equal to v = sqrt(2gh/3), where g = 9.8 m/s^2 is the acceleration of free fall.
By purchasing this digital product, you receive a ready-made solution to the problem that can be used for educational purposes. The solution is written by an experienced author and contains detailed explanations for each step of the solution.
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Solution of problem D1-20 V.A. Dievsky is a determination of the speed of a parachutist at the moment of landing during a vertical descent without an initial speed from a height of 200 meters, taking into account the presence of air resistance force, which is proportional to the square of the speed and has a value of R = 3mv^2.
To solve the problem, it is necessary to use the equation of body motion taking into account the force of air resistance. The equation will look like:
mg - R = ma
where m is the mass of the parachutist, g is the acceleration of gravity, R is the force of air resistance, a is the acceleration of the parachutist.
It is also necessary to use the equation for the air resistance force, which is proportional to the square of the speed:
R = k*v^2
where k is the proportionality coefficient, v is the parachutist’s speed.
Substituting the expression for R into the equation of motion, we get:
mg - kv^2 = m*a
To solve the problem it is necessary to find the speed v at the moment of landing. To do this, you can use the law of conservation of energy:
mgh = (1/2)mv^2
where h is the initial descent height.
From this equation we can express the speed v:
v = sqrt(2gh)
Substituting this expression for v into the equation of motion, we obtain:
mg - k(2gh) = m*a
Where can we express acceleration a:
a = g - (2kg*h)/m
Thus, the speed of the parachutist at the moment of landing will be equal to:
v = sqrt(2gh) = sqrt(29.81200) ≈ 198.26 m/s
The acceleration of the parachutist at the moment of landing will be equal to:
a = g - (2kgh)/m = 9.81 - (23v^2)/(m9.81) ≈ 8.16 m/s^2
Answer: the parachutist’s speed at the moment of landing is approximately 198.26 m/s, acceleration is about 8.16 m/s^2.
The solution to problem D1 option 20 (D1-20) is a textbook created by the author Dievsky V.A. and intended for students preparing to take the mathematics exam. The manual contains a detailed solution to problem D1 option 20, which is included in the list of exam tasks. The author provides readers with a complete analysis of the problem, examines its conditions step by step, gives recommendations and tips to help them understand the material and successfully solve the problem. The publication can be useful both for independent study of the material and for use by teachers as an additional teaching aid for students.
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