Question 1190706
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A Ferris wheel that is 18 meter in diameter makes a revolution in 5 seconds. 
If the center is 10m above the ground, how long does a rider reach a height of 15 m?
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<pre>
Our Ferris wheel has the radius of 18/2 = 9 meters and the lowest position at the height of 1 meter over the ground.


Let's consider this Ferris wheel as rotated anti-clockwise.



So, it rotates from position 6 pm (the hour hand) anti-clockwise to the position 3 pm by 90 degrees,

reaching the height of 10 meters above the ground, and then rotates for some additional angle "a"

anti-clockwise, such that the height is 15-10 = 5 meters above the level of the center, which is 10 meters above the ground.



This angle "a" satisfies equation  sin(a) = {{{5/9}}} = 0.555556,  so  a = arcsin(0.555556) = 33.75 degrees.


Thus the wheel turns/rotates the angle  90 + 33.75 degrees = 123.75 degrees, total.


The angular velocity of the wheel is 360 degrees in 5 seconds, or  {{{360/5}}} = 72 degrees per second.


So, it will take  {{{123.75/72}}} = 1.72 seconds for the rider to get the height of 15 meters above the ground,
starting from the lowest position.


<U>ANSWER</U>.  1.72 seconds, approximately.
</pre>

Solved.