A person standing close to the edge on top of a 96-foot building throws a ball vertically upward. The quadratic function h = − 16 t 2 + 116 t + 96 models the ball's height above the ground, h , in feet, t seconds after it was thrown. a) What is the maximum height of the ball? b) How many seconds does it take until the ball hits the ground?

 a)  306.25 feet   b) 8 s

Step-by-step explanation:

Actually we have to find the function' s  h(t)  maximum meaning.

To do that we have to find the corresponding t - let call it t max

As known t max= (t1+t2)/2 where t1 and t2 are the roots of quadratic equation' s  

Lets find the roots t1 and t2

-16*t^2+116*t+96=0   divide by 4 each side of the equation

-4*t^2 +29*t+24=0

D=29^2+24*4*4=1225 =35^2

t1=(-29-35)/(-8)=8

t2=(-29+35)/(-8)=-6/8=-3/4=-0.75

t max=  (8+(-0.75))= 7,25/2=3.625 s

h max= -16*t max ^2+116*t +96= -16*3.625^2+116*3.625+96=306.25 feet

b) t2=8s is the time when the ball hits the ground.

  a) 306.25 ft

  b) 8 seconds

Step-by-step explanation:

a) The time at the maximum height is found from the equation for the axis of symmetry:

  ax^2 +bx +c   has axis of symmetry at x=-b/(2a)

For the given equation, the t-value at the vertex is ...

  t = -116/(2(-16)) = 3.625 . . . seconds

At that time, the height is ...

  h = (-16(3.625) +116)(3.625) +96 = (58)(3.625) +96 = 306.25

The maximum height is 306.25 feet.

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b) The ball will hit the ground when h=0. From the vertex values in the first part, we know we can rewrite the equation in vertex form as ...

  h(t) = -16(t -3.625)^2 +306.25

This will be 0 when ...

  0 = -16(t -3.625)^2 +306.25

  (t -3.625)^2 = 306.25/16

  t = 3.625 +√19.140625 = 3.625 +4.375 = 8

The ball will hit the ground after 8 seconds.

astronomers is your final answer! : )