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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/s2, how high above the ground is the ball 2 s after it is thrown?

Respuesta :

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the asnwer is 12 feet

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Step-by-step explanation:

It is given that, two boys are throwing a baseball back and forth.

Height above ground, h₀ = 4 ft

Initial velocity of child, u = 36 ft/s

Acceleration due to gravity, a = -16 ft/s²

The equation of projectile motion is given by :

[tex]h(t)=-16t^2+ut+h_o[/tex]

According to the given condition,  

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

h(t) = 2.356 meters

Height of the ball at t = 2 s

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

h = 12 ft

So, at t = 2 seconds the height of the ball is 12 feet from the ground. Hence, this is the required solution.

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