I have this problem with equation h=15+20t-2t^2 object throwen from tower

I must making graph to show time of flight
Then show greatest height reached.
How high is building
And how long did it take for object to hit.

1 I thinking for graph formula must be xv=-b/2a
which is -20/-2x2=-20/-4=5 this is for my x.
can you please showing me how for getting my y .

when i am knowing this i think i can do rest of problems,

Thanksing you

2. 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.

3. Hello, gregorio!

Equation: .$\displaystyle h\:=\:15+20t-2t^2$, object thrown from tower

I must make a graph to show time of flight.
(a) Then show greatest height reached.
(b) How high is building?
(c) How long did it take for object to hit?

We have a parabola: .$\displaystyle h \:=\:-2t^2 + 20t + 15$
. . So: .$\displaystyle a = -2,\;b = 20,\;c = 15$

Then the vertex is at: .$\displaystyle t \:=\:\frac{-b}{2a} \:=\:\frac{-20}{2(-2)} \:=\:5$
. .
It reaches maximum height in 5 seconds.

Hence: .$\displaystyle h \:=\:-2(5^2) + 20(5) + 15 \:=\:65$
. .
The maximum height is 65 feet.

(b) How high is the building?
Hint: Where was the object at the very beginning $\displaystyle (t = 0)$?

(c) When does it strike the ground?
Hint: When does the object have zero height $\displaystyle (h = 0)$?