Two stones are thrown up from the ground, one stone is 3 times faster than the other. The faster stone takes 10 seconds to return to the ground so the slower stone takes 3.3 seconds to return. If the slower stone reaches a max height of H, how high in terms of H will the faster stone go? I don't understand how to show this part....

Question

Two stones are thrown up from the ground, one stone is 3 times faster than the other.  The faster stone takes 10 seconds to return to the ground so the slower stone takes 3.3 seconds to return.  If the slower stone reaches a max height of H, how high in terms of H will the faster stone go?  I don't understand how to show this part....

anonymous · Answer

So you know that the initial speeds of the stones differ by a factor of 3

anonymous · Answer

You can calculate the max height of the faster stone without worrying about the times.

anonymous · Answer

One of the standard equations for motion under uniform acceleration doesn't include the time - can you state it ?

anonymous · Answer

I believe the equation I want is: $$H = y _{0} + v^{2}_{yo}/2g$$  I'm not sure how to apply it to this problem....

anonymous · Answer

yes i think that is correct, and you can just set Yo equal to zero since the stones start out from ground level

anonymous · Answer

you can write out the same equation for each stone separately

anonymous · Answer

you know H for the slow stone, and you know the ratio of their speeds
that allows you then to find H for the faster stone

anonymous · Answer

you need to combine those two equations somehow