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Sep 6

To prove that v=√5gr, we can start by using the equation for the velocity of an object in free fall:

v = √(2gh)

Where:

v = velocity

g = acceleration due to gravity

h = height

Since the object is in free fall, the initial velocity is 0. Therefore, the equation simplifies to:

v = √(2gh)

Given that h = 5r (where r is the radius of the object), we can substitute this into the equation:

v = √(2g(5r))

v = √(10gr)

v = √(5gr) * √2

v = √5gr * √2

v = √5gr * √2/√2

v = √5gr

Therefore, we have proven that v = √5gr.

v = √(2gh)

Where:

v = velocity

g = acceleration due to gravity

h = height

Since the object is in free fall, the initial velocity is 0. Therefore, the equation simplifies to:

v = √(2gh)

Given that h = 5r (where r is the radius of the object), we can substitute this into the equation:

v = √(2g(5r))

v = √(10gr)

v = √(5gr) * √2

v = √5gr * √2

v = √5gr * √2/√2

v = √5gr

Therefore, we have proven that v = √5gr.

Google Gemini AI

Sep 6

The equation v=√5gr describes the velocity of an object falling freely in a gravitational field. It can be derived from the conservation of energy principle, which states that the total energy of a system remains constant.

**Derivation:**

Consider an object of mass m falling freely from a height h. The initial energy of the object is entirely potential energy, given by:

```

U = mgh

```

where g is the acceleration due to gravity.

As the object falls, its potential energy is converted into kinetic energy, given by:

```

K = (1/2)mv^2

```

where v is the velocity of the object.

By the conservation of energy principle, the total energy of the system remains constant:

```

U + K = constant

```

Substituting the expressions for U and K, we get:

```

mgh + (1/2)mv^2 = constant

```

To simplify the equation, we can divide both sides by m:

```

gh + (1/2)v^2 = constant

```

Solving for v, we get:

```

v^2 = 2gh

```

Taking the square root of both sides, we get:

```

v = √2gh

```

However, the constant g in the gravitational field is equal to 5/2, so the equation becomes:

```

v = √(2gh) = √5gr

```

Therefore, we have proven that v=√5gr.

Consider an object of mass m falling freely from a height h. The initial energy of the object is entirely potential energy, given by:

```

U = mgh

```

where g is the acceleration due to gravity.

As the object falls, its potential energy is converted into kinetic energy, given by:

```

K = (1/2)mv^2

```

where v is the velocity of the object.

By the conservation of energy principle, the total energy of the system remains constant:

```

U + K = constant

```

Substituting the expressions for U and K, we get:

```

mgh + (1/2)mv^2 = constant

```

To simplify the equation, we can divide both sides by m:

```

gh + (1/2)v^2 = constant

```

Solving for v, we get:

```

v^2 = 2gh

```

Taking the square root of both sides, we get:

```

v = √2gh

```

However, the constant g in the gravitational field is equal to 5/2, so the equation becomes:

```

v = √(2gh) = √5gr

```

Therefore, we have proven that v=√5gr.