Three equations of motion homework

[Viraam]

Homework Statement

Homework Equations

$2as = v^2-u^2$
$v = 0 ... \text{at maximum height}$
$\therefore s = \frac{-(u)^2}{2a} \\ a = -g \\ \therefore s = \frac{-(u)^2}{-2g} = \frac{(u)^2}{2g}$
Please explain using only the Three equations of motion and not anything related to Energy or other formulae.

The Attempt at a Solution

Is my answer correct:-

 

[Doc Al]

You are using the kinematic equations for uniformly accelerated motion. Do you think they would apply?

 

[Viraam]

You are using the kinematic equations for uniformly accelerated motion. Do you think they would apply?

Yeah. Because they are uniformly accelerated to $g$

 

[Doc Al]

Yeah. Because they are uniformly accelerated to $g$

The acceleration is $g$ when the balls are close to the Earth's surface. But those speeds are pretty high.

 

[Viraam]

The acceleration is $g$ when the balls are close to the Earth's surface. But those speeds are pretty high.

Ohh then please tell me how to find the Heights.

 

[Doc Al]

Use conservation of energy.

 

[Viraam]

Use conservation of energy.

We haven't learned that. Can you please tell me the formula.
Some other sites states the following:-
$KE + PE \text{ at the surface} = PE \text{ at the maximum height}$
Can you please explain in simpler words because I am very new to this.

 

[Doc Al]

We haven't learned that. Can you please tell me the formula.
Some other sites states the following:-
$KE + PE \text{ at the surface} = PE \text{ at the maximum height}$
Can you please explain in simpler words because I am very new to this.

That's exactly the equation you need. Of course you need to learn how to express the gravitational PE.

If you haven't studied this yet, where did you get the problem?

 

[Viraam]

We haven't learned it completely yet. But this question was in our chapter called Gravitation.

 

[Doc Al]

We haven't learned it completely yet. But this question was in our chapter called Gravitation.

Good. So it should cover how you calculate the gravitational PE at some distance from a planet.

 

[Viraam]

No we learned only these equations:-
$F = \frac{GMm}{R^2}\\ g = \frac{GM}{R^2}$
Do you mean this:-
$PE = mgh$
$PE = \frac{GMmh}{R^2}$

 

[Doc Al]

The equation I mean is this one:

Read about it here: Gravitational Potential Energy

 

[Viraam]

The equation I mean is this one:
View attachment 88836

Read about it here: Gravitational Potential Energy

Thank You
So the answer is 4:1

 

[Viraam]

The equation I mean is this one:
View attachment 88836

Read about it here: Gravitational Potential Energy

I have one doubt: When do we use Gravitational Potential Energy?

 

[Doc Al]

So the answer is 4:1

How did you determine that?

 

[Viraam]

How did you determine that?

Using
$KE + PE \text{ at surface} = PE \text{ at maximum height}$
$\frac{1}{2} m *(2\sqrt{\frac{gR}{3}})^2 - \frac{GMm}{R}= \frac{-GMm}{R+h}\\ h = 2 R\\ \text{For the other ball:-} \\ h' = \frac{R}{2}\\ \frac{h}{h'} = \fbox{4:1}\\$

 

[haruspex]

Using
$KE + PE \text{ at surface} = PE \text{ at maximum height}$
$\frac{1}{2} m *(2\sqrt{\frac{gR}{3}})^2 - \frac{GMm}{R}= \frac{-GMm}{R+h}\\ h = 2 R\\ \text{For the other ball:-} \\ h' = \frac{R}{2}\\ \frac{h}{h'} = \fbox{4:1}\\$

Looks right to me. Well done.

 

[Doc Al]

Using
$KE + PE \text{ at surface} = PE \text{ at maximum height}$
$\frac{1}{2} m *(2\sqrt{\frac{gR}{3}})^2 - \frac{GMm}{R}= \frac{-GMm}{R+h}\\ h = 2 R\\ \text{For the other ball:-} \\ h' = \frac{R}{2}\\ \frac{h}{h'} = \fbox{4:1}\\$

Good work! (Just wanted to check. )

 

[Viraam]

Good work! (Just wanted to check. )

Have a doubt. How exactly do we know when to use Gravitational Potential Energy. Please explain in the simplest words. Thanking you in advance.

 

[Doc Al]

Have a doubt. How exactly do we know when to use Gravitational Potential Energy. Please explain in the simplest words.

That's difficult to answer. Just stating the obvious: When gravity is involved and a massive object changes position, you may be able to make use of gravitational PE. But it depends on the problem and what you need to find.

 

[Viraam]

That's difficult to answer. Just stating the obvious: When gravity is involved and a massive object changes position, you may be able to make use of gravitational PE. But it depends on the problem and what you need to find.

Ohh ok Thank you so much