An object is propelled vertically upward from the top of a 256-foot building. The quadratic function s(t) = -16t squared + 192t + 256 models the ball's height above the ground, s(t), in feet, t seconds after it was thrown. How many seconds does it take until the object finally hits the ground? Round to the nearest tenth of a second if necessary. A. 6 seconds

t= 13.2 seconds

Step-by-step explanation:

The quadratic function for ball's height in terms of time t is given as

s(t)= -16t² +192t + 256

Now we want to find out the time at which the ball hits the ground.

When the ball hits the ground, the height of the ball will become zero so in the above equation we can put s(t)=0

0= -16t² +192t + 256

or 16t² -192t - 256 = 0

Solving this quadratic equation, we have

t= 13.211 and t= -1.211

Since time can't be negative so

t= 13.2 seconds

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

can you put the picture up?

step-by-step explanation: