Answer:
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
