Question

Камень бросают вертикально вверх с начальной скоростью 6 м/с. На какой высоте скорость камня уменьшиться в двое?

Answer 1

v₁=6 м/с

v₂=3 м/с

h₂=0 м

h₁-?

Тут нужно помнить закон сохранения энергии

E(k1)+E(n2)=E(k2)+E(n1)

mv²₁/2 + mgh₂=mv²₂/2 + mgh₁  ,      h₂=0 м  ⇒

mv²₁/2=mv²₂/2 + mgh₁

mv²₁/2 - mv²₂/2=mgh₁

(v²₁-v²₂)/2=gh₁

h₁=(v²₁-v²₂)/2g=((6м/с)²-(3м/с)²)/2×10м/с²=1,35 м

Ответ:1,35 м

Answer 2

Calculating the Height at Which the Stone's Velocity is Halved

To calculate the height at which the stone's velocity is halved, we can use the following formula:

v^2 = u^2 - 2gh

Where: - v = final velocity (when the stone's velocity is halved, it becomes 3 m/s as the initial velocity is 6 m/s) - u = initial velocity (6 m/s) - g = acceleration due to gravity (approximately 9.81 m/s^2) - h = height

We can rearrange the formula to solve for h:

h = (u^2 - v^2) / (2g)

Substitute the given values into the formula to find the height at which the stone's velocity is halved.

h = (6^2 - 3^2) / (2 * 9.81)

h ≈ 1.53 meters

So, the stone's velocity will be halved at a height of approximately 1.53 meters.