Question

4. Тіло кидають вертикально вгору зі швидкістю 25 м/с. На якій висоті кінетична енергія тіла вдвічі більша, ніж його потенціальна енергія? Опором повітря знехтуйте.

Answer 1

Даний: початкова швидкість тіла при підніманні вгору v = 25 м/с.

Формула для потенціальної енергії тіла в полі тяжіння:

E_p = mgh

Формула для кінетичної енергії тіла:

E_k = (mv^2)/2

При підніманні тіла вгору механічна енергія (сума кінетичної та потенціальної енергій) зберігається, тобто:

E_p + E_k = const

Оскільки швидкість тіла на максимальній висоті дорівнює нулю, то кінетична енергія на цій висоті дорівнює нулю, тому:

E_p = const

При дослідженні питання про те, на якій висоті кінетична енергія тіла вдвічі більша, ніж потенціальна енергія, можна записати рівняння:

E_k = 2E_p

(mv^2)/2 = 2mgh

Звідси:

h = v^2/(2g*2) = v^2/8g

Підставляємо дані:

h = (25 м/с)^2 / (8 * 9.81 м/с²) ≈ 6.36 м

Отже, на висоті близько 6.36 м кінетична енергія тіла вдвічі більша, ніж потенціальна енергія.

Answer 2

Question Analysis

The question asks at what height the kinetic energy of an object thrown vertically upward with a speed of 25 m/s is twice its potential energy. The question specifies that air resistance should be ignored.

Answer

To find the height at which the kinetic energy of the object is twice its potential energy, we can use the conservation of mechanical energy. The total mechanical energy of the object is the sum of its kinetic energy and potential energy.

Let's assume the initial kinetic energy of the object is K and the initial potential energy is P. At the highest point of its trajectory, the object will have zero kinetic energy and maximum potential energy. Therefore, the total mechanical energy at the highest point is equal to the initial potential energy.

According to the conservation of mechanical energy, the total mechanical energy remains constant throughout the motion of the object, as long as no external forces (such as air resistance) are acting on it.

Mathematically, we can express this as:

K + P = P

Since we want to find the height at which the kinetic energy is twice the potential energy, we can rewrite the equation as:

2K = P

Now, let's solve for the height.

Calculation

The kinetic energy of an object is given by the formula:

K = (1/2)mv^2

where m is the mass of the object and v is its velocity.

In this case, the object is thrown vertically upward, so its velocity changes sign at the highest point. Therefore, we need to consider the magnitude of the velocity, which is 25 m/s.

The potential energy of an object at a certain height h is given by the formula:

P = mgh

where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height.

Now, let's substitute the values into the equation:

2K = P

2 * (1/2)mv^2 = mgh

Simplifying the equation:

v^2 = 2gh

Substituting the given velocity (v = 25 m/s) into the equation:

(25 m/s)^2 = 2 * 9.8 m/s^2 * h

Simplifying further:

625 m^2/s^2 = 19.6 m/s^2 * h

Solving for h:

h = 625 m^2/s^2 / (19.6 m/s^2)

h ≈ 31.89 meters

Therefore, the height at which the kinetic energy of the object is twice its potential energy is approximately 31.89 meters.

Conclusion

The height at which the kinetic energy of the object is twice its potential energy, when a body is thrown vertically upward with a speed of 25 m/s and air resistance is ignored, is approximately 31.89 meters.