Question

Задача. Камень брошен со скоростью 30 м/с под углом 60(градусов) к горизонту. Каковы скорость и ускорение камня в верхней точке траектории? Будьте добрый расписать всё как можно подробнее.

Answer 1

Поехали. Начальная скорость камня (в проекциях)

$v_{0x} = v_0\cos\alpha\\ v_{0y} = v_0\sin\alpha$

Где v0 - модуль начальной скорости, альфа - угол броска

Теперь законы изменения скорости

$v_x(t) = v_{0x}\\ v_y(t) = v_{0y}-gt$

Верхняя точка характеризуется тем, что в ней игрек-проекция скорости равна нулю (тело не летит ни вверх ни вниз). Значит в верхней точке остается только икс-проекция, то есть в верхней точке скорость равна

$v = v_x(t_1) = v_{0x} = v_0\cos\alpha = v_0/2$

где  t1 - время подъема.

Ускорение тела в верхней точке такое же как и в любой другой точке и равняется g

Answer 2

Problem:

We are given that a stone is thrown with a velocity of 30 m/s at an angle of 60 degrees to the horizontal. We need to find the velocity and acceleration of the stone at the highest point of its trajectory.

Solution:

To solve this problem, we can break down the initial velocity of the stone into its horizontal and vertical components. Then, we can analyze the motion of the stone at the highest point of its trajectory.

Let's start by finding the horizontal and vertical components of the initial velocity.

The horizontal component of the initial velocity can be found using the formula:

Horizontal component of velocity = Initial velocity * cos(angle)

Substituting the given values, we have:

Horizontal component of velocity = 30 m/s * cos(60 degrees)

To find the vertical component of the initial velocity, we can use the formula:

Vertical component of velocity = Initial velocity * sin(angle)

Substituting the given values, we have:

Vertical component of velocity = 30 m/s * sin(60 degrees)

Now, let's calculate the horizontal and vertical components of the initial velocity.

Horizontal component of velocity: Horizontal component of velocity = 30 m/s * cos(60 degrees) = 30 m/s * 0.5 = 15 m/s.

Vertical component of velocity: Vertical component of velocity = 30 m/s * sin(60 degrees) = 30 m/s * 0.866 = 25.98 m/s.

At the highest point of the trajectory, the vertical component of the velocity becomes zero. This is because the stone reaches its maximum height and starts to fall back down.

Now, let's analyze the motion of the stone at the highest point of its trajectory.

At the highest point, the stone's velocity is purely horizontal, and its acceleration is purely vertical. The acceleration due to gravity acts vertically downward and has a magnitude of 9.8 m/s^2.

Therefore, at the highest point of the trajectory: - The velocity of the stone is equal to the horizontal component of the initial velocity, which is 15 m/s. - The acceleration of the stone is equal to the acceleration due to gravity, which is 9.8 m/s^2.

To summarize: - The velocity of the stone at the highest point of its trajectory is 15 m/s. - The acceleration of the stone at the highest point of its trajectory is 9.8 m/s^2.

I hope this helps! Let me know if you have any further questions.