Help with mechanics

A ball at rest of dropped at height h m onto a horizontal surface. After striking the surface it rebounds to half its original height. Show that the time
t taken from the instance the ball is dropped to the instant it strikes the floor for the second time is given by t=(2+sqrt2)(sqrt (h/g)).
Where g is the acceleration due to gravity

Should be a relatively straightforward application of suvat? What are you stuck with?

I just thought we don’t have all the necessary info for suvat. S=h, u=0 is all I got

You can find the time to hit the ground in terms of h and g?
Thats the first part.

How?

What suvat relates u, a, t,S

Would it be something like this? I’m having trouble making t the subject

why u=30? The ball is at rest initially.

Sorry, I’m getting confused between questions

This?

What will u be then bcs it won’t be zero?

Are you talking about the upwards or diwnwards phase of the bounce?

I did this for the whole thing from the initial drop to the second time it strikes the ground

The rebound complicates things? You can't do a single calculation.

What values am I using for the rebound. I know h and g, but what is the third one?