Aball is kicked upward with an initial velocity of 56 feet per second. the ball's height, h (in feet), from the ground is modeled by -16t^2+56t , where t is measured in seconds. what is the practical domain in this situation?

Domain in this situation is [ 3.5 , ∞ )

Step-by-step explanation:

Given:

Function is -16t² + 56t

This function gives the height, h of ball at time t.

So, h(t) =  -16t² + 56t

To find : Domain of the function h(x)

We know that height of the ball can not be negative. So, put function greater than equal to 0.

⇒ h(t) ≥ 0

⇒ -16t² + 56t  ≥ 0

-8t ( 2t - 7 ) ≥ 0

⇒ -8t ≥ 0  

   t ≥ 0

and 2t - 7 ≥ 0

2t ≥ 7

t ≥ 7/2

t ≥ 3.5

So, Domain of the function = { x : x ∈ R and x ≥ 3.5 } = [ 3.5 , ∞ )

Therefore, Domain in this situation is [ 3.5 , ∞ )

three-fourths of an hour

36÷48=0.75

0.75=