## Answer :

[tex] H(t) = -16t^{2} + 36t + 4 [/tex]

[tex] H(2) = -16(2)^2 + 36(2) + 4 [/tex]

[tex]H(2) = -64 + 72 + 4 [/tex]

[tex]H(2) = 12 [/tex]

Therefore, two seconds after being thrown the ball is 12 feet above the ground.

**Answer:**

Two seconds after being thrown the ball is 12 feet above the ground.

**Step-by-step explanation:**

**Given value,**

Height = 4 ft

Time t = 2 sec

Velocity u = 36 ft/s

Using equation of motion.

[tex]s=ut-\dfrac{1}{2}at^2+h_{0}[/tex]....(I)

Where, u = velocity

h = height

t = time

a = acceleration

Put the value in the equation (I)

[tex]s=36\times2-\dfrac{1}{2}\times32\times4+4[/tex]

[tex]s =12\ m[/tex]

**Hence, Two seconds after being thrown the ball is 12 feet above the ground.**