Projectile Motion, Find Gravity

[fgc_grapplerGOD]

Homework Statement

You kick a ball with a speed of 2 m/s, at a 45 degree inclination to the horizontal. You measure h to be 4⁄15 m. What planet are you on?

Homework Equations

The Attempt at a Solution

Upon first glance, I thought that this problem did not provide enough information. However, my instructor insisted that it does, so I re-examined. My next idea was to take the following two equations:
0 = v_{0}Sin\Theta -gt
\frac{4}{15} = v_{0}Sin\Theta - \frac{1}{2}gt^{2}
From the first equation, I found:
t = \frac{v_{0}sin\Theta }{g}
Then I substituted this into the second equation. After some manipulation, I ended up with:
\frac{4}{15} = v_{0}sin\Theta -\frac{1}{2}\frac{v_{0}^{2}sin^{2}\Theta }{g}
Which simplifies to:
v_{0}^{2}sin^{2}\Theta = (v_{0}sin\Theta - \frac{4}{15})2g
Finally:
g = \frac{v_{0}^{2}sin^{2}\Theta }{2(v_{0}sin\Theta - \frac{4}{15})}
At this point I would simply substitute my known values. However, I admit that I am not entirely confident in this answer so I wanted to see if there is perhaps and easier approach, or if my approach is even close at all. Thank you.

 

[SammyS]

Homework Statement

You kick a ball with a speed of 2 m/s, at a 45 degree inclination to the horizontal. You measure h to be 4⁄15 m. What planet are you on?
[ IMG]http://i.imgur.com/D2SIr4d.jpg[/PLAIN]

Homework Equations

The Attempt at a Solution

Upon first glance, I thought that this problem did not provide enough information. However, my instructor insisted that it does, so I re-examined. My next idea was to take the following two equations:
0 = v_{0}Sin\Theta -gt
\frac{4}{15} = v_{0}Sin\Theta - \frac{1}{2}gt^{2}
From the first equation, I found:
t = \frac{v_{0}sin\Theta }{g}
Then I substituted this into the second equation. After some manipulation, I ended up with:
\frac{4}{15} = v_{0}sin\Theta -\frac{1}{2}\frac{v_{0}^{2}sin^{2}\Theta }{g}
Which simplifies to:
v_{0}^{2}sin^{2}\Theta = (v_{0}sin\Theta - \frac{4}{15})2g
Finally:
g = \frac{v_{0}^{2}sin^{2}\Theta }{2(v_{0}sin\Theta - \frac{4}{15})}
At this point I would simply substitute my known values. However, I admit that I am not entirely confident in this answer so I wanted to see if there is perhaps and easier approach, or if my approach is even close at all. Thank you.

Hello fgc_grapplerGOD. Welcome to PF.

Your second equation is missing a $\ t \ .$

$\displaystyle \ \Delta s=v_0\,t+(1/2)a\,t^2\$

 

[fgc_grapplerGOD]

Hello fgc_grapplerGOD. Welcome to PF.

Your second equation is missing a $\ t \ .$

$\displaystyle \ \Delta s=v_0\,t+(1/2)a\,t^2\$

Ah thank you very much, rookie mistake there :P.
So after making the necessary adjustments I have:
\frac{4}{15} = (v_{0}sin\Theta )(\frac{v_{0}sin\Theta }{g})-\frac{1}{2g}(\frac{v_{0}sin\Theta }{g})^{2}
Which simplifies to:
\frac{4}{15} = \frac{v_{0}^{2}sin^{2}\Theta }{g} -\frac{1}{2}\frac{v_{0}^{2}sin^{2}\Theta }{g}
Finally, I found:
g = \frac{15}{4}(v_{0}^{2}sin^{2}\Theta -\frac{1}{2}v_{0}^{2}sin^{2}\Theta )
At which point I would just substitute my known values. This approach is sufficient?

 

[SammyS]

Ah thank you very much, rookie mistake there :P.
So after making the necessary adjustments I have:
\frac{4}{15} = (v_{0}sin\Theta )(\frac{v_{0}sin\Theta }{g})-\frac{1}{2g}(\frac{v_{0}sin\Theta }{g})^{2}
...

What is g doing in the denominator with the 2 from the 1/2 ?

... but it looks like the final result you gave is fine.

 

[fgc_grapplerGOD]

What is g doing in the denominator with the 2 from the 1/2 ?

... but it looks like the final result you gave is fine.

That was just an error I made in converting from my paper solution into Latex. Thank you for your help!

 

[SammyS]

That was just an error I made in converting from my paper solution into Latex. Thank you for your help!

Of course you can simplify that result.

 

[fgc_grapplerGOD]

Of course you can simplify that result.

Yes I actually just did that. I simplified to:
g = \frac{v_{0}^{2}sin^{2}\theta}{2h}
Which is actually a much cleaner result that I had anticipated. Of course my final answer was:
\frac{15}{4} \frac{m}{s^{2}}
After doing some research we can conclude the projectile was on either Mars or Mercury.