Question

Камень, брошенный вертикально вверх с поверхности Земли с некоторой скоростью, упал на Землю через 2 с. Через какое время упадет камень, брошенный вертикально вверх с той же скоростью на Луне? Ускорение свободного падения на Луне в 6 раз меньше, чем на Земле. Распишите решение, пожалуйста.

Answer 1

Законы физики связаны с ускорением свободного падения.
F = M*a
РЕШЕНИЕ
a = g/6 = 1/6*g
Время полета по формуле
T =  V/a = V/(1/6*g) = 6*2 = 12 c - ОТВЕТ

Answer 2

Problem Analysis

We are given that a stone is thrown vertically upwards from the surface of the Earth with a certain velocity and it takes 2 seconds to fall back to the Earth. We need to determine how long it will take for a stone thrown vertically upwards with the same velocity to fall back to the Moon. The acceleration due to gravity on the Moon is 6 times smaller than on Earth.

Solution

To solve this problem, we can use the equations of motion for vertical motion under constant acceleration. The key equation we will use is:

h = V0t - (1/2)gt^2

where: - h is the height of the stone above the surface of the Earth or Moon, - V0 is the initial velocity of the stone, - g is the acceleration due to gravity on the Earth or Moon, and - t is the time taken for the stone to reach the height h.

Let's assume the initial velocity of the stone thrown vertically upwards on both the Earth and the Moon is V0. We know that the acceleration due to gravity on the Moon is 6 times smaller than on Earth. Therefore, the acceleration due to gravity on the Moon (g_moon) can be calculated as:

g_moon = (1/6)g_earth

where g_earth is the acceleration due to gravity on Earth.

We are given that it takes 2 seconds for the stone to fall back to the Earth. Using this information, we can calculate the initial velocity of the stone on Earth (V0_earth) using the equation:

h_earth = V0_earth * t_earth - (1/2)g_earth * t_earth^2

where h_earth is the height of the stone above the surface of the Earth, t_earth is the time taken for the stone to reach the height h_earth, and g_earth is the acceleration due to gravity on Earth.

Now, we need to find the time taken for the stone to fall back to the Moon (t_moon). We can use the equation:

h_moon = V0_moon * t_moon - (1/2)g_moon * t_moon^2

where h_moon is the height of the stone above the surface of the Moon, t_moon is the time taken for the stone to reach the height h_moon, V0_moon is the initial velocity of the stone on the Moon, and g_moon is the acceleration due to gravity on the Moon.

To find t_moon, we can rearrange the equation as follows:

t_moon = (V0_moon + sqrt(V0_moon^2 + 2g_moon * h_moon)) / g_moon

Substituting the values of g_moon and h_moon, we can calculate t_moon.

Let's calculate the values step by step.

Calculation

Given: - t_earth = 2 seconds - g_moon = (1/6)g_earth

From the equation h_earth = V0_earth * t_earth - (1/2)g_earth * t_earth^2, we can calculate V0_earth.

From the equation t_moon = (V0_moon + sqrt(V0_moon^2 + 2g_moon * h_moon)) / g_moon, we can calculate t_moon.

Solution

Using the given information and equations, we can calculate the time taken for the stone to fall back to the Moon.

Let's calculate step by step:

Step 1: Calculate V0_earth using the equation h_earth = V0_earth * t_earth - (1/2)g_earth * t_earth^2.

Step 2: Calculate g_moon using the equation g_moon = (1/6)g_earth.

Step 3: Calculate t_moon using the equation t_moon = (V0_moon + sqrt(V0_moon^2 + 2g_moon * h_moon)) / g_moon.

Step 1: Calculate V0_earth

We are given that t_earth = 2 seconds. Let's assume g_earth = 9.8 m/s^2 (acceleration due to gravity on Earth).

Using the equation h_earth = V0_earth * t_earth - (1/2)g_earth * t_earth^2, we can calculate V0_earth.

Substituting the values, we have:

0 = V0_earth * 2 - (1/2) * 9.8 * 2^2

Simplifying the equation, we get:

0 = 2V0_earth - 19.6

2V0_earth = 19.6

V0_earth = 19.6 / 2

V0_earth = 9.8 m/s

Therefore, the initial velocity of the stone thrown vertically upwards on Earth is V0_earth = 9.8 m/s.

Step 2: Calculate g_moon

We are given that g_moon = (1/6)g_earth.

Substituting the value of g_earth = 9.8 m/s^2, we have:

g_moon = (1/6) * 9.8

g_moon = 1.6333 m/s^2

Therefore, the acceleration due to gravity on the Moon is g_moon = 1.6333 m/s^2.

Step 3: Calculate t_moon

We need to find the time taken for the stone to fall back to the Moon (t_moon).

Using the equation t_moon = (V0_moon + sqrt(V0_moon^2 + 2g_moon * h_moon)) / g_moon, we can calculate t_moon.

Since the stone is thrown vertically upwards on the Moon with the same initial velocity as on Earth (V0_moon = V0_earth = 9.8 m/s), we can substitute the values:

t_moon = (9.8 + sqrt(9.8^2 + 2 * 1.6333 * h_moon)) / 1.6333

Now, we need to find the value of h_moon. We know that the stone takes 2 seconds to fall back to the Earth. Using this information, we can calculate h_earth using the equation h_earth = V0_earth * t_earth - (1/2)g_earth * t_earth^2.

Substituting the values, we have:

h_earth = 9.8 * 2 - (1/2) * 9.8 * 2^2

Simplifying the equation, we get:

h_earth = 19.6 - 19.6

h_earth = 0 m

Therefore, the height of the stone above the surface of the Earth is h_earth = 0 m.

Now, substituting the value of h_earth = 0 m in the equation for t_moon, we have:

t_moon = (9.8 + sqrt(9.8^2 + 2 * 1.6333 * 0)) / 1.6333

Simplifying the equation, we get:

t_moon = (9.8 + sqrt(9.8^2)) / 1.6333

t_moon = (9.8 + 9.8) / 1.6333

t_moon = 19.6 / 1.6333

t_moon ≈ 12 seconds

Therefore, the time taken for the stone to fall back to the Moon is approximately 12 seconds.

Answer

The stone, thrown vertically upwards with the same velocity, will take approximately 12 seconds to fall back to the Moon.

Conclusion

In this problem, we calculated the time taken for a stone thrown vertically upwards with a certain velocity to fall back to the Moon. We used the equations of motion for vertical motion under constant acceleration and the given information about the acceleration due to gravity on the Moon. The stone takes approximately 12 seconds to fall back to the Moon.