Asoccer ball is kicked into the air from a height of 3 feet with an initial velocity of 72 feet per second. how many seconds is the ball 84 feet above the ground?

For 2.25 seconds the ball is 84 feet above the ground.

Step-by-step explanation:

Initial height when the ball is kicked = 3 feet

Initial velocity of the soccer ball = 72 feet per second

The height(h) of the vertical thrown object is modeled by the formula :

h(t) = -16t² + vt + s , where t is the time, v is initial velocity and s is initial height

Now, h(t) is given to be 84 feet and we need to find the time taken by the ball to reach 84 feet above the ground :

84 = -16t² + 72t + 3

⇒ 16t² - 72t + 81 = 0

Finding roots of this equation :

Hence, for 2.25 seconds the ball is 84 feet above the ground.

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2