An object is thrown off a 256-foot-tall building, and the distance of the object from the ground is measured every second. The function that models the height, h, of the object after t seconds is h(t) = –16t2 + 96t + 256. Determine the time when the object hits the ground

The object hits the ground 8 seconds after launch.

Step-by-step explanation:

Let , where is the height of the object above the ground, measured in feet, and is time, measured in seconds. We notice that given expression is a second-order polynomial and hence, there are two roots, one of them corresponding to the instant when the object hits the ground. If , we get the following expression:

(Eq. 1)

Roots are found by applying the Quadratic Formula:

or

Only the first root offer a solution that is physically reasonable. Thus, the object hits the ground 8 seconds after launch.

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐