Question

A seat on a ferris wheel is level with the center of the wheel. the diameter of the wheel is 210 feet (ft). if the wheel rotates 45° counterclockwise and stops, how much does the height of the seat change? startfraction 210 startroot 2 endroot over 2 endfraction ft startfraction 105 startroot 2 endroot over 2 endfraction ft 52.5 ft 26.25 ft

Answer 1

The 210 feet diameter of the ferris wheel rotated 45° gives;

The amount by which the height of the seat changes is the option;

• startfraction 210 startroot 2 endroot over 2 endfraction ft.

$\Delta y = \frac{210 \cdot \sqrt{2}}{2} \: ft$

Which method can be used to find the change in height of the ferris wheel?

Given parameters;

Diameter of the wheel, D = 210 feet

Angle of rotation of the wheel = 45°

Required;

The measure by which the height changes.

Solution;

The equation of a Ferris wheel can be represented as follows;

y = A•sin(B•(x - c)) + D

Change in height, ∆y, is found as follows;

∆y = A•sin(t1) + D - A•sin(t2) + D

∆y = A•(sin(t1) - sin(t2))

Where;

t1 = 0

t2 = 45°

A = The radius = 210 feet

Therefore;

∆y = 210×2×(sin((45+0)/2)° × cos((45-0)/2)°)

∆y = 210×2×(sin(22.5)° × cos(22.5)°)

∆y = 210×sin(2×22.5)° = 210×sin(45)°

• ∆y = 210×(√2)/2

The amount by which the height of the seat changes is therefore;

$\Delta y = \frac{210 \cdot \sqrt{2}}{2} \: ft$

The correct option is therefore;

startfraction 210 startroot 2 endroot over 2 endfraction ft.

Learn more about the equation of a Ferris wheel here:

brainly.com/question/14504727

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