Question

Тело свободно падает с небольшой высоты на планете с ускорением 2 м / С2, начальная скорость 0 м / с, который час будет на 3 метре

Answer 1

Начало координат в точке падения, ось координат направлена вертикально вниз.

$h=\frac{a t^2}{2}$
$t=\sqrt{\frac{2 h}{a}}=\sqrt{\frac{2 \cdot 3}{2}} \approx 1,7(c)$

Answer 2

H=Vot+gt²/2
Vо=0, g =2м/c², h=3 м
t²=2h/g
t=√(2h/g)
t=√(2*3/2)=√3 ≈ 1,73 c

Answer 3

Calculation of the time it takes for an object to fall from a height of 3 meters on a planet with an acceleration of 2 m/s^2

To calculate the time it takes for an object to fall from a height of 3 meters on a planet with an acceleration of 2 m/s^2, we can use the equation of motion for free fall:

h = (1/2)gt^2 + v0t + h0

Where: - h is the height (3 meters in this case) - g is the acceleration due to gravity (2 m/s^2 in this case) - t is the time - v0 is the initial velocity (0 m/s in this case) - h0 is the initial height (0 meters in this case)

We can rearrange the equation to solve for t:

t = sqrt((2h)/g)

Substituting the given values:

t = sqrt((2 * 3) / 2) = sqrt(3) seconds

Therefore, it will take approximately 1.73 seconds for the object to fall from a height of 3 meters on the planet with an acceleration of 2 m/s^2.