Question

Помогите решить, очень прошу.Тело подброшено вверх со скоростью 20м/с с балкона высотой в 25м. Каково его перемещение за 2; 4; 5 сек? ускорение Ах = -10м/с в квадрате.если можен, пожалуйста развернутый ответ, как записать...

Answer 1

S=Vo*t+ Подставляем:S1=20*2-20м S2=20*4-80=0  S3=-20*5+125+25

Answer 2

Problem Statement

You are given the following information: - An object is thrown upwards from a balcony with an initial velocity of 20 m/s. - The height of the balcony is 25 m. - The object experiences an acceleration of -10 m/s^2 in the horizontal direction.

You need to find the displacement of the object after 2 seconds, 4 seconds, and 5 seconds.

Solution

To solve this problem, we need to consider the motion of the object in the vertical direction and the horizontal direction separately.

# Vertical Motion

In the vertical direction, the object is thrown upwards with an initial velocity of 20 m/s. The acceleration due to gravity is acting in the opposite direction, causing the object to slow down and eventually reverse its direction. The height of the balcony is 25 m.

To find the displacement of the object after a certain time, we can use the following kinematic equation:

Displacement (vertical) = Initial velocity * time + (1/2) * acceleration * time^2

Substituting the given values: - Initial velocity = 20 m/s (upwards) - Acceleration = -10 m/s^2 (opposite to the direction of motion) - Time = 2 s, 4 s, 5 s

Let's calculate the displacement for each time interval:

For 2 seconds: Displacement (vertical) = 20 * 2 + (1/2) * (-10) * (2^2) = 40 - 20 = 20 m For 4 seconds: Displacement (vertical) = 20 * 4 + (1/2) * (-10) * (4^2) = 80 - 80 = 0 m For 5 seconds: Displacement (vertical) = 20 * 5 + (1/2) * (-10) * (5^2) = 100 - 125 = -25 m

# Horizontal Motion

In the horizontal direction, the object experiences a constant acceleration of -10 m/s^2. This means that its velocity decreases by 10 m/s every second.

Since the acceleration is constant, we can use the following kinematic equation to find the displacement:

Displacement (horizontal) = Initial velocity * time + (1/2) * acceleration * time^2

Substituting the given values: - Initial velocity = 0 m/s (since the object is thrown vertically) - Acceleration = -10 m/s^2 (constant) - Time = 2 s, 4 s, 5 s

Let's calculate the displacement for each time interval:

For 2 seconds: Displacement (horizontal) = 0 * 2 + (1/2) * (-10) * (2^2) = 0 - 20 = -20 m For 4 seconds: Displacement (horizontal) = 0 * 4 + (1/2) * (-10) * (4^2) = 0 - 80 = -80 m For 5 seconds: Displacement (horizontal) = 0 * 5 + (1/2) * (-10) * (5^2) = 0 - 125 = -125 m

Summary

The displacement of the object after 2 seconds is 20 m upwards and 20 m horizontally to the left. The displacement of the object after 4 seconds is 0 m vertically and 80 m horizontally to the left. The displacement of the object after 5 seconds is -25 m downwards and 125 m horizontally to the left.

Please let me know if you need any further assistance!