Find the time of its fall

You have the correct equation. All you need to do is set T=1 and you know that the distance travelled is 0.5665h. Then you can solve for the height put it back in the original equation and solve for total time.

 

you would have [tex] 0.5665 h =\frac{1}{2}g[/tex]

Now rearrange for h.

When you find h put it back into the original equation and solve for t.

 

No worries its my fault. Give me a minute (doing many things at once) and I'll correct my mistake.

 

Ok so the equation that describes the motion is:

[tex] s=h-\frac{1}{2}gt^2 [/tex]

Now the question tells you that in the final second the particle falls [itex]s=0.5665h [/itex] therefore,

[tex] 0.5665h = h-v_t-\frac{1}{2}g [/tex]

where v_t is the velocity at t-1 seconds. You can work out this velocity and should get a quadratic in h.

 

What about other kinematic equations? Do you know any more?

 

There are some basic formula here

https://www.physicsforums.com/showthread.php?t=110015