Umm, you are using formulas to find the vertex of the parabola of the flight of the object.

y = ax^2 +bx +c -------(i), parabola

x of vertex = -b/(2a)

y of vertex = c -(b^2)/(4a)

So,

h = 15 +20t -2t^2 ------parabola

Rearrange that to be in the form of (i) above,

h = -2t^2 +20t +15 ----------(ii)

t of vertex = "-b/(2a)" = -20/(2(-2)) = 5 ------which you got.

h of vertex = "c -(b^2)/(4a) = 15 -[(20^2)/(4(-2))] = 15 +50 = 65

Therefore, greatest height reached is 65 units.

How high is the building?

When time t is zero, or when ball is not thrown yet,

h = 15 +20t -2t^2

h = 15 +20(0) -2(0)^2

h = 15 units ---------------height of building.

And how long did it take for object to hit.

To hit the ground?

That will be if h = -15 units. (The negative sign is to show only that the distance is measured opposite the positive measurements.)

Because the gr4ound is 15 units below the top of the building---where t begins.

So,

-15 = 15 +20t -2t^2

Put them all to the lefthand side,

2t^2 -20t -15 -15 = 0

2t^2 -20t -30 = 0

t^2 -10t -15 = 0

Umm, can you take over now?

Use the Quadratic formula to find t when the object hit the ground. It is t = 11.3245 units time.

If you like, check that against h = 15 +20t -2t^2. See if you you can get h = -15.