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.
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}...
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 EquationsThe Attempt at a Solution
Upon first glance, I thought that this problem did not provide enough information...