An iron ball is bobbing up and down on the end of a spring. The maximum height of the ball is 46 inches and its minimum height is 18 inches. It takes the ball 2 seconds to go from its maximum height to its minimum height. Which model best represents the height, h, of the ball after t seconds?

Respuesta :

Answer:

The equation for height is

[tex]h=32sin(\frac{\pi}{2}t)+14[/tex]

Step-by-step explanation:

we are given

An iron ball is bobbing up and down on the end of a spring

So, height function must be trigonometric in nature

So, we can use formula

[tex]h=Asin(Bt)+D[/tex]

now, we can find A , B and D

Calculation of A:

maximum height 46 inch

minimum height =18 inch

so,

[tex]A=\frac{46+18}{2}=32[/tex]

Calculation of B:

It takes the ball 2 seconds to go from its maximum height to its minimum height

So, half of time period is 2 sec

[tex]\frac{T}{2}=2[/tex]

[tex]T=4[/tex]

now, we can use period formula

[tex]T=\frac{2\pi}{B}[/tex]

we can find B

[tex]4=\frac{2\pi}{B}[/tex]

[tex]B=\frac{2\pi}{4}[/tex]

[tex]B=\frac{\pi}{2}[/tex]

Calculation of D:

Max=46

min=18

[tex]D=\frac{46-18}{2}[/tex]

[tex]D=14[/tex]

now, we can plug these values into formula

and we get

[tex]h=32sin(\frac{\pi}{2}t)+14[/tex]


Answer:

Required model is  [tex]h=32\sin(\frac{\pi}{2}t)+14[/tex]

Step-by-step explanation:

Given : An iron ball is bobbing up and down on the end of a spring. The maximum height of the ball is 46 inches and its minimum height is 18 inches. It takes the ball 2 seconds to go from its maximum height to its minimum height.

To find : Which model best represents the height, h, of the ball after t seconds?

Solution :  

According to question,

The height function must be trigonometric in nature.

So, we can use formula,

[tex]h=A\sin(Bt)+D[/tex]

Now, We calculate A,B and D

1) Maximum height 46 inch

Minimum height =18 inch

Average height is A

[tex]A=\frac{46+18}{2}=32[/tex]

2) It takes the ball 2 seconds to go from its maximum height to its minimum height

So, half of time period is 2 sec

i.e, [tex]\frac{T}{2}=2[/tex]

[tex]T=4[/tex]

Period is B

[tex]T=\frac{2\pi}{B}[/tex]

[tex]4=\frac{2\pi}{B}[/tex]

[tex]B=\frac{2\pi}{4}[/tex]

[tex]B=\frac{\pi}{2}[/tex]

3) D is the midline

So, Max=46  and min=18

[tex]D=\frac{46-18}{2}[/tex]

[tex]D=14[/tex]

Substituting all the values,

A=32 , D=14 , [tex]B=\frac{\pi}{2}[/tex]

[tex]h=A\sin(Bt)+D[/tex]

[tex]h=32\sin(\frac{\pi}{2}t)+14[/tex]

Therefore, Required model is  [tex]h=32\sin(\frac{\pi}{2}t)+14[/tex]