Question

тело кинули вертикально вверх со скоростью 40 м на секунду сейчас подъёма на максимум высоты и скорость и время действий​

Answer 1

Ответ:

По з-ну сохранения энергии: Вся кинетическая энергия перейдет в потенциальную

$\frac{ (mv^2)}{2} = mgh$

$h=\frac{v^2}{2g}$

$h = \frac{1600}{20} = 80$ м

Теперь найдем время через ур-ие кинематики:

$S = v_{0} t +\frac{at^2}{2}$ (оно в векторном виде)

Спроецируем его на ось Oy, направленную вертикально вверх:

$h=v_{0} t-\frac{gt^2}{2}$

Домножим ур-ние на 2

$2h=2v_{0} t-gt^2$

Подставим известные величины:

$2*80=2*40t-10t^2$

$160=80t-10t^2$

Поделим все на 10:

$16=8t-t^2$

Перенесем все в одну сторону:

$t^2-8t+16=0$

Решим квадратное уравнение относительно t, заметив что это полный квадрат:

$t^2-8t+16=(t-4)^2$

$(t-4)^2=0$

$t-4=0$

$t=4$ c

Ответ: h=80м

            t=4c

Объяснение:

Answer 2

Problem Analysis

We are given that an object is thrown vertically upwards with an initial velocity of 40 m/s. We need to determine the maximum height reached by the object, as well as the time it takes to reach that height.

Solution

To solve this problem, we can use the equations of motion for uniformly accelerated motion. Since the object is thrown vertically upwards, we can assume that the acceleration due to gravity is acting in the opposite direction to the motion of the object.

Let's break down the problem into two parts: the upward motion and the downward motion.

Upward Motion

During the upward motion, the object is slowing down due to the acceleration due to gravity. At the maximum height, the object momentarily comes to rest before starting its downward motion.

We can use the following equation to find the time taken to reach the maximum height:

v = u + at

Where: - v is the final velocity (0 m/s at the maximum height) - u is the initial velocity (40 m/s) - a is the acceleration due to gravity (-9.8 m/s^2) - t is the time taken to reach the maximum height

Rearranging the equation, we get:

t = (v - u) / a

Substituting the values, we have:

t = (0 - 40) / (-9.8) = 4.08 seconds

So, it takes approximately 4.08 seconds for the object to reach the maximum height.

To find the maximum height reached by the object, we can use the equation:

s = ut + (1/2)at^2

Where: - s is the displacement (maximum height) - u is the initial velocity (40 m/s) - t is the time taken to reach the maximum height (4.08 seconds) - a is the acceleration due to gravity (-9.8 m/s^2)

Substituting the values, we have:

s = (40)(4.08) + (1/2)(-9.8)(4.08)^2 = 83.04 meters

Therefore, the maximum height reached by the object is approximately 83.04 meters.

Summary

- The time taken to reach the maximum height is approximately 4.08 seconds. - The maximum height reached by the object is approximately 83.04 meters.

Please let me know if I can help you with anything else.