Melody

can't you do ANY of your own thinking?

I am with Rom, I think that + is meant to be a comma.  

Once you respond to our query we might be able to answer you.

Yes I forgot the 0.

No big deal, I showed you how to do it. 

You can pick up trivial errors like that on your own.

You do not need to be sorry - I did it correctly. 

A thank you would be nice though.

Hi Rom, I agreed with you when i first read the question but we interpreted it differently from intended.

This is a simplified school question.  Perhaps we are intended to use calculus, or perhaps not.

I will do it without calculus.

The trajectory of a ball will be parabolic (pre-knowledge on my part).  We are only interested in height and time.

The x does not refer to horizonal distance it refers to time and I am going to change it to a t make it less confusing.

So the ball is traveling in the path of a concave down parabola.   (t,y)     (time, height)

If you let the top of the building be the point (0,0)    that is 0 time and  0 vertical height 

then the ground will be ( 5,-25)

and the maximum height coordinates will be (2,20)

It is a concave down parabola and the t intercepts will be 0 and  4    (becasue it is symmetrical)

so the eqution must be

\(y=-a(t-0)(t-4)\\ y=-at(t-4)\\ When \;\;t=2,\;y=20\\ sub\\ 20=-a*2(2-4)\\ 20=-2a*-2\\ 20=4a\\ a=5\\ \text{So if the coordinates are set as I have described, I mean with the top of the buildng being height 0}\\ then\\~\\ y=-5t(t-4) \text{ Will model the path of the ball} \)

I hope that helps.

I already answered you (A full day before you edited your question) 

You have not responded to my answer in any way.

What is wrong with you?  

Are you just totally rude?

I even read your mind and answered the question that you wanted to ask, rather than the garbled nonsense that you wrote.

There seems to be 2 different questions here.

I will answer the first one.

Let  a  and  b  be real numbers. The complex number    4 - 5i     is a root of the quadratic

\(z^2 + (a + 8i) z + (-39 + bi) = 0\).

I am told that one solution is z=4-5i

so

\((4-5i)^2 + (a + 8i) (4-5i) + (-39 + bi) = 0\\ \text{expand and simplify and we get}\\ (4a-8)+(b-5a-8)i=0\\ so\\ 4a-8=0\\ a=2\\ b-5a-8=0\\ b=18\\ \text{so the equation becomes}\\ z^2 + (2 + 8i) z+ (-39 + 18i) = 0\\ \)

Now I will use the quadratic formula to solve this.

\(z^2 + (2 + 8i) z+ (-39 + 18i) = 0\\ z = {-(2 + 8i) \pm \sqrt{\triangle} \over 2}\\ (-1-4i)\pm\frac{\sqrt{\triangle}}{2}=4-5i \qquad \text{I'm not sure if it is + or - so i will keep it together}\\ \pm\frac{\sqrt{\triangle}}{2}=5-i\\ \text{So the solutions are}\\ z=-1-4i\pm (5-i)\\ z=-1-4i+ (5-i) \qquad z=-1-4i- (5-i)\\ z=4-5i \qquad \qquad \quad \qquad z=-6-3i\\~\\ \text{So the other root is } -6-3i\)

\(d\; \alpha\; t^2\\ d=kt^2\\ find\;k\\ 80=k*4^2\\ k=5\\ d=5t^2\\ after\;12\;sec\\ d=5*12^2\\ d=720m\\ 720-80=640 \)

So in the next 8 sec the ball will drop 640m

Thanks x-squared :)

Of course, how silly of me, I should have thought of that!

I got a different answer from you but I did not get any sqrt6   

so I suppose the c is 0