# Calculating acceleration due to gravity on an alien planet

1. Homework Statement
On an alien planet you throw a stone upward and measure the times Ta and Tb. The difference in heights of point a and point b is h. Calculate the acceleration due to gravity.
Ta = 7.1s; Tb= 4.4s; h = 15.1m

2. Homework Equations
a = dv/st
v(t) = vo - gt

3. The Attempt at a Solution
I have tried solving a = t*(2gh)^.5. I have also tried a = 2h/(Ta-Tb). I keep getting numbers that are far too great I believe based off of the numbers I am given.

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Hootenanny
Staff Emeritus
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1. Homework Statement
On an alien planet you throw a stone upward and measure the times Ta and Tb. The difference in heights of point a and point b is h. Calculate the acceleration due to gravity.
Ta = 7.1s; Tb= 4.4s; h = 15.1m

2. Homework Equations
a = dv/st
v(t) = vo - gt

3. The Attempt at a Solution
I have tried solving a = t*(2gh)^.5. I have also tried a = 2h/(Ta-Tb). I keep getting numbers that are far too great I believe based off of the numbers I am given.
I'm afraid that you cannot use either of those two equations, since you do not know that initial and final velocities of the stone. And what is that g doing in your attempt at the solution? Here g and a are the same thing.

Can you think of a better kinematic (SUVAT) equation to use rather than the one that you have chosen?

I don't understand how you are suppose to use kinematic equations when you do not have any of the variable to work with.

Hootenanny
Staff Emeritus
Gold Member
I don't understand how you are suppose to use kinematic equations when you do not have any of the variable to work with.
Well that not quite right, we do no some things. Let's make two lists. First, write down that variable that you want to find out. Second, write down a list of everything that you know.

okay so we have two time variables, a height displacement and we want to find acceleration. But most equations ask for velocity, which I do not know. So do I need to solve for the velocity at each time given. Then find the difference between those velocities and then divided that by the difference in time?

Hootenanny
Staff Emeritus
Gold Member
okay so we have two time variables, a height displacement and we want to find acceleration.
Good. So now list the (four) kinematic equations for constant acceleration.

v=at+vo
dv=.5(v+vo)
x=.5at^2 + vot + xo
x=dvt + xo = (v^2 - vo^2)/2a + xo

Hootenanny
Staff Emeritus
Gold Member
v=at+vo
dv=.5(v+vo)
x=.5at^2 + vot + xo
x=dvt + xo = (v^2 - vo^2)/2a + xo
Good. So, we know x, t and v1. We want to find out a, but we don't know v0.

Can you use two of the above equations to eliminate v0?

I get: .5at^2 + xo = x

Hootenanny
Staff Emeritus
Gold Member
the only question i have left is: x is equal to h and i assume xo=0 right?

another question: for the time is it the time difference?

Hootenanny
Staff Emeritus
Gold Member
another question: for the time is it the time difference?
Correct.
the only question i have left is: x is equal to h and i assume xo=0 right?
Yup okay so with numbers i get: (2*15.1)/((7.1-4.4)^2) = 4.143 which is not correct. what am i doing wrong?

okay I think I am doing something wrong since I keep getting vt=0.

Hootenanny
Staff Emeritus
Gold Member
okay so with numbers i get: (2*15.1)/((7.1-4.4)^2) = 4.143 which is not correct. what am i doing wrong?
I don't see any reason why your answer should be wrong, except perhaps that it may need a negative sign since the acceleration is vertically downwards.

i have tried -4.143 m/s^2 and it is still wrong

Hootenanny
Staff Emeritus
Gold Member
i have tried -4.143 m/s^2 and it is still wrong
Does the online submission program specify a required accuracy?

i did finally get the answer. Thank you for all your help.

Hootenanny
Staff Emeritus
My pleasure 