Question

Тело падает с некоторой высоты с начальной скоростью 2 м/с. Найти высоту, с которой упало тело на землю, если движение длилось 3с.

Answer 1

Дано:
V0=2 м/с
t=3 c
H-?
Решение:
$H= V_{0} + \frac{gt^2}{2} = 2+ \frac{10*9}{2} =2+45=47$
Ответ: 47 м/с

Answer 2

Calculation of the Height of the Fall

To calculate the height from which an object fell when it has an initial velocity and the time of motion, we can use the equations of motion. In this case, the object falls with an initial velocity of 2 m/s and the motion lasts for 3 seconds.

The equation that relates the height (h), initial velocity (v), and time (t) for an object in free fall is:

h = v * t + (1/2) * g * t^2

Where: - h is the height - v is the initial velocity - t is the time - g is the acceleration due to gravity (approximately 9.8 m/s^2)

Substituting the given values into the equation, we can calculate the height:

h = 2 * 3 + (1/2) * 9.8 * 3^2

Simplifying the equation:

h = 6 + (1/2) * 9.8 * 9

h = 6 + 44.1

h = 50.1 meters

Therefore, the object fell from a height of approximately 50.1 meters.