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Two boys are throwing a baseball back and forth. The ball is 4 ft above the ground when it leaves one child’s hand with an upward velocity of 36 ft/s. If acceleration due to gravity is –16 ft/s^2, how high above the ground is the ball 2 s after it is thrown?
h(t) = at2 + vt + h0

Sagot :

chipperrider
[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.

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.

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