Question

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?

Answer 1

a) 306.25 ft

b) 8 seconds

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.

Answer 2

Answer: a) 306.25 feet b) 8 s

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.