Circle question

[sadpwner]

A ferris wheel as a diameter of 24 metres and completes one revolution every 48 seconds. The centre of the wheel is 18 metres off the ground. The car is directly 12m beneath the centre. How high above the ground, in metres, will the car be 18 seconds later.
Original answer is 18+6*square root 2

I keep getting 24 as an answer don't know why it has the extra square root.

so (18/48)*100= percentage traveled=37.5
37.5%*360=angel traveled=135
We already know that from the ground to the centre it is 18m so
135-90=45
Then i got every 1m height is (24*pi)/2/24 of the circumference
so then i got (24*pi*45/360)/(24*pi)/2/24 +18=24

somehow it is wrong

can anyone help me from here?

 

[Deleted member 4993]

A ferris wheel as a diameter of 24 metres and completes one revolution every 48 seconds. The centre of the wheel is 18 metres off the ground. The car is directly 12m beneath the centre. How high above the ground, in metres, will the car be 18 seconds later.
Original answer is 18+6*square root 2

I keep getting 24 as an answer don't know why it has the extra square root.

so (18/48)*100= percentage traveled=37.5
37.5%*360=angel traveled=135
We already know that from the ground to the centre it is 18m so
135-90=45
Then i got every 1m height is (24*pi)/2/24 of the circumference
so then i got (24*pi*45/360)/(24*pi)/2/24 +18=24

I do not understand this part of your logic.

somehow it is wrong

can anyone help me from here?

You are correct that the car has travelled 135° from vertical position.

Now consider the position of the car relative to the horizontal line through the center of the wheel.

The car is at a height of 12 * sin (45°) off from this line.

so height of the car - from the ground

= height of the horizontal line through the center of the wheel + height of the car from this line

= 18 + 12 * sin (45°) = 18 + 12 * 1/√2 = 18 + 12 * √2/2 .................and continue....

 

[srmichael]

A ferris wheel as a diameter of 24 metres and completes one revolution every 48 seconds. The centre of the wheel is 18 metres off the ground. The car is directly 12m beneath the centre. How high above the ground, in metres, will the car be 18 seconds later.
Original answer is 18+6*square root 2

I keep getting 24 as an answer don't know why it has the extra square root.

so (18/48)*100= percentage traveled=37.5
37.5%*360=angel traveled=135
We already know that from the ground to the centre it is 18m so
135-90=45
Then i got every 1m height is (24*pi)/2/24 of the circumference
so then i got (24*pi*45/360)/(24*pi)/2/24 +18=24

somehow it is wrong

can anyone help me from here?

Well, I tried to draw a picture, but I couldn't get it to show up in my reply.

See if you can follow what my picture was going to show.

After 18 seconds you are correct in that it will have gone 135° from its starting position and thus from the center to its position it is at a 45° angle. You now have a 45-45-90 triangle with the center as one vertex and the ferris wheel cage as another vertex. The distance from the center to the cage is the hypotenuse and is also the radius or 12 m. You are thus needing one leg of this triangle which is the additional height above the already 18 m that the center of the ferris wheel is located. Using the ratio of sides for a 45-45-90 triangle this additional distance would be 6√2. So the total height of the cage is 18 + 6√2 m.

I hope you could follow that.

P.S. Subhotosh Khan's explanation will work for any angle relative to the center so make sure you understand his method as well as you will not always have the cage at 45° from the center.