Acceleration of gravity NOT ON EARTH

[Justhelp]

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

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

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

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

 

[SammyS]

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

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

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

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 ?

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

 

[Justhelp]

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

21x6=126 final velocity right?

 

[Justhelp]

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

And thanks for the welcome :)

 

[SammyS]

21x6=126 final velocity right?

No.

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

 

[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^.

 

[Justhelp]

And I came up with 3.5 ms^.

 

[berkeman]

Use the equations of motion for a constant acceleration (in this case, the acceleration of gravity on that planet). See the Description of Motion in One Dimension here:

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#motcon

.

 

[berkeman]

And I came up with 3.5 ms^.

I'm getting a slightly different answer. Can you post your work?

 

[HallsofIvy]

21x6=126 final velocity right?

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.

 

[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 canceled on the right.

126/36=a

 

[Justhelp]

Hallsofivy what is itex?

 

[SammyS]

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 canceled on the right.

126/36=a

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.

 

[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^

 

[Justhelp]

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

 

[HallsofIvy]

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!

 

[Justhelp]

I mean, meters per second sqaured.

 

[Justhelp]

9.8 being the gravity on earth..

 

[HallsofIvy]

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

 

[Justhelp]

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

 

[Justhelp]

I felt like a ref. point would help me remember that a stood for gravity.

 

[azizlwl]

equation for constant acceleration
s=ut+(1/2)at²
s=126m
t=6sec
a=?m/s² - acceleration due to gravity on planet whoosedat
u=?m/s - initial velocity.

Here 2 unknowns for one equation. No unique answer.

 

[Ratch]

Justhelp,

This is a differential equation problem. If the acceleration is constant, then there are cut and dried formulas that can be used. The attachment shows how differential methods are employed. The final velocity is 42 meters/second and the average velocity is 21 meters/second.

Ratch