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ChatGPT
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.
0
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.
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.
0
