Question

Из поднимающегося вертикально вверх вертолета со скоростью u на высоте H вылетает тело. Через сколько времени оно упадет на Землю? Какой будет скорость у тела? Сопротивления воздуха нет.

Answer 1

V=U-gt - равноускоренное движение
mV^2/2+0=mU^2/2+mgH - закон сохранения енергии
V^2=U^2+2gH (1) - нашли скорость
t=(U-V)/g, подставим скорость V из (1) и найдем время
дальше все арифметика

Answer 2

Problem Analysis

We are given a scenario where a body is ejected from a helicopter that is ascending vertically upwards with a speed of 'u' at a height 'H'. We need to determine the time it takes for the body to fall to the ground and the velocity of the body when it hits the ground. We are also told that there is no air resistance.

Solution

To solve this problem, we can use the equations of motion under constant acceleration. Since the body is falling freely under the influence of gravity, we can assume the acceleration due to gravity, 'g', to be constant.

Let's break down the problem into two parts: 1. Time taken for the body to fall to the ground. 2. Velocity of the body when it hits the ground.

Part 1: Time taken for the body to fall to the ground

When the body is ejected from the helicopter, it will have an initial vertical velocity equal to the upward velocity of the helicopter, 'u'. As the body falls freely under the influence of gravity, its velocity will increase due to the constant acceleration of gravity.

To find the time taken for the body to fall to the ground, we can use the equation of motion:

H = ut + (1/2)gt^2

Where: - H is the initial height of the body (given) - u is the initial vertical velocity of the body (equal to the upward velocity of the helicopter) - g is the acceleration due to gravity (approximately 9.8 m/s^2) - t is the time taken for the body to fall to the ground (unknown)

We can rearrange the equation to solve for 't':

(1/2)gt^2 + ut - H = 0

This is a quadratic equation in 't'. We can solve it using the quadratic formula:

t = (-u ± √(u^2 + 2gH)) / g

Since we are interested in the positive value of 't' (time cannot be negative), we can ignore the negative solution.

Now let's substitute the given values into the equation and calculate the time taken for the body to fall to the ground.

Part 2: Velocity of the body when it hits the ground

To find the velocity of the body when it hits the ground, we can use the equation of motion:

v = u + gt

Where: - v is the final velocity of the body (unknown) - u is the initial vertical velocity of the body (equal to the upward velocity of the helicopter) - g is the acceleration due to gravity (approximately 9.8 m/s^2) - t is the time taken for the body to fall to the ground (calculated in Part 1)

Now let's substitute the values of 'u' and 't' into the equation and calculate the velocity of the body when it hits the ground.

Calculation

Given: - Initial upward velocity of the helicopter, u = [u] - Height from which the body is ejected, H = [H] - Acceleration due to gravity, g = 9.8 m/s^2

Using the equation for time taken for the body to fall to the ground:

t = (-u ± √(u^2 + 2gH)) / g

Substituting the given values:

t = (-[u] ± √(([u])^2 + 2 * 9.8 * [H])) / 9.8

Now, let's calculate the time taken for the body to fall to the ground.

Using the equation for the velocity of the body when it hits the ground:

v = u + gt

Substituting the given values:

v = [u] + 9.8 * t

Now, let's calculate the velocity of the body when it hits the ground.

Answer

The time taken for the body to fall to the ground is [calculated time] seconds. The velocity of the body when it hits the ground is [calculated velocity] m/s.

Please note that the values of 'u' and 'H' need to be provided in order to calculate the time and velocity accurately.