# Acceleration of gravity NOT ON EARTH

1. Jun 11, 2012

### Justhelp

This was the first day of class, and this is due Wed. Thank you.

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

An object undergoes free fall 126 m on the planet whoosedat. It takes 6.00 sec. to travel the 126 m what is the acceleration of gravity on whoosedat?

2. Relevant equations

How long would it take an object to fall 15.0 m on whoosedat?

3. The attempt at a solution

1. It is distance/time? making it 21m/s

2. 21m/s/15? making it 1.4 sec

2. Jun 11, 2012

### SammyS

Staff Emeritus
Hello Justhelp. Welcome to PF !

That gives the average velocity for the 6 seconds.

If the object started from rest, what is its final velocity, if its average velocity is 21 m/s ?

3. Jun 11, 2012

### Justhelp

21x6=126 final velocity right?

4. Jun 11, 2012

### Justhelp

Is that what I have to do before I reach the results? I am not seeing it.

And thanks for the welcome :)

5. Jun 11, 2012

### SammyS

Staff Emeritus
No.

For constant acceleration, $\displaystyle v_{\text{Average}}=\frac{v_{\text{Iniial}}+v_{ \text{Final}}}{2}\ .$ ... just an average of the two.

6. Jun 11, 2012

### Justhelp

My dad was just explaining to me per second per second.. And I couldn't quite get it. Could you explain? I understand that acceleration has to always have ms^.

7. Jun 11, 2012

### Justhelp

And I came up with 3.5 ms^.

8. Jun 11, 2012

### Staff: Mentor

9. Jun 11, 2012

### Staff: Mentor

10. Jun 11, 2012

### HallsofIvy

Staff Emeritus
The problem tells you that the object falls 126 meters. That is a distance, not a velocity. Assuming that the distance is short enough that acceleration can be taken to be a constant, g, then the change in speed over t seconds is gt and the distance fallen is $(g/2)t^2$ (assuming that the initial speed was 0). With t= 6 seconds, you have $(g/2)6^2= 126$. Solve that for g and then complete the problem.

11. Jun 11, 2012

### Justhelp

a=9.8 m/s^

d= a (m(s^)

126= a (m(s^)

126=a(m(6^)
126= a(m(36)
note: 36 is divided and cancelled on the right.

126/36=a

12. Jun 11, 2012

### Justhelp

Hallsofivy what is itex?

13. Jun 11, 2012

### SammyS

Staff Emeritus
Why are you using g for Earth, 9.8 m/s2, all of a sudden?

If the object starts from rest, accelerates uniformly and has an average velocity of 126m/s, then it's final velocity (at the end of 6 seconds) must have been 252 m/s.

14. Jun 11, 2012

### Justhelp

Because the problem I have come to conclude is not giving the gravity, it may be implying it but I can not tell, the implied gravity in general is 9.8/m s^

15. Jun 11, 2012

### Justhelp

Thanks for fixing that Hallsofivy :). Your solution is very much similar to mine. :) it helps!

16. Jun 11, 2012

### HallsofIvy

Staff Emeritus
You keep writing "/ms^". What in the world does that mean? Speed or velocity is "distance divided by time", m/s or "meters per second". Acceleration is speed divided by seconds and so "(meters per second) per second", m/s^2.

"the implied gravity in general is 9.8/m s^". Where is that implied? The whole point of the problem is to calculate the acceleration due to gravity from [/itex](a/2)(6^2)= 126[/itex]. Simple algebra tells you that a is NOT 9.8!

17. Jun 11, 2012

### Justhelp

I mean, meters per second sqaured.

18. Jun 11, 2012

### Justhelp

9.8 being the gravity on earth..

19. Jun 11, 2012

### HallsofIvy

Staff Emeritus
Yet, the whole point of the problem is that this is NOT on the earth so that is completely irrelevant.

20. Jun 11, 2012

### Justhelp

a is the gravity # 9.8 is just a ref. I didn't use 9.8 at all in the problem.