Answer :
[tex]H(t) = at^{2} + vt + H_{0} [/tex]
[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.
[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.