Homework Help: Three equations of motion homework

1. Sep 18, 2015

Viraam

1. The problem statement, all variables and given/known data

2. Relevant 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.
3. The attempt at a solution

2. Sep 18, 2015

Staff: Mentor

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

3. Sep 18, 2015

Viraam

Yeah. Because they are uniformly accelerated to $g$

4. Sep 18, 2015

Staff: Mentor

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

5. Sep 18, 2015

Viraam

Ohh then please tell me how to find the Heights.

6. Sep 18, 2015

Staff: Mentor

Use conservation of energy.

7. Sep 18, 2015

Viraam

We haven't learnt 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.

8. Sep 18, 2015

Staff: Mentor

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?

9. Sep 18, 2015

Viraam

We havent learnt it completely yet. But this question was in our chapter called Gravitation.

10. Sep 18, 2015

Staff: Mentor

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

11. Sep 18, 2015

Viraam

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

12. Sep 18, 2015

Staff: Mentor

13. Sep 18, 2015

Viraam

14. Sep 18, 2015

Viraam

15. Sep 18, 2015

Staff: Mentor

How did you determine that?

16. Sep 18, 2015

Viraam

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

17. Sep 19, 2015

haruspex

Looks right to me. Well done.

18. Sep 19, 2015

Staff: Mentor

Good work! (Just wanted to check. )

19. Sep 19, 2015

Viraam

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

20. Sep 19, 2015

Staff: Mentor

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.