The length of time it takes an object to fall to the ground varies directly as the square root of the height from which is realesed. if an object released from a height 100 feet hits the ground after 2.5 seconds, find a modeling equation gives times t in seconds for an object released from height h feet

a. The modeling equation is t = 0.25√h

b. The object released from height 282.24 feet

Step-by-step explanation:

* Lets explain how to solve the problem

- Direct variation is a relationship between two variables that can

 be expressed by an equation in which one variable is equal to a

 constant times the other

- Ex: If y varies directly with x , y ∝ x , then y = k x, where k is the

 constant of variation

- The value of k is founded by the initial values of x and y

* Lets solve the problem

- The length of time it takes an object to fall to the ground varies

 directly as the square root of the height from which is released

∴ t ∝ √h, where t is the length of the time in second and h is the

  height in feet

∴ t = k √h , where k is the constant of variation

a.

- An object released from a height 100 feet hits the ground after

 2.5 seconds

∴ h = 100 feet

∴ t = 2.5 seconds

∵ t = k√h

∴ 2.5 = k√100

∴ 2.5 = k(10)

- Divide both sides by 10

∴ k = 0.25

∴ t = 0.25 h

* The modeling equation is t = 0.25√h

b.

- An object falls for 4.2 seconds

∴ t = 4.2 seconds

∵ t = 0.25√h

∴ 4.2 = 0.25 √h

- Divide both sides by 0.25

∴ 16.8 = √h

- To find h square the both sides

∴ h = 282.24

* The object released from height 282.24 feet

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

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step-by-step explanation: