sadpwner wrote:A ferris wheel has a diameter of 24 metres and completes one revolution every 48 seconds. The centre of the wheel is 18 metres of the ground. The car is directly 12 metres beneath the ferris wheel centre. The car is 6 metres above the air. How high above the ground will the car be if it moved for 18 seconds.

I know where the car will be but i don't know how to get the height of where it lands. So i just did (24*pi/48)*18. Also i figgured out that the car rotated around the circle for 135 degrees. 18/48 *360

I don't understand. How do you know "where the car will be" without knowing its height, etc? What do you mean by "doing" the number you show?

Try using what they give you. Draw a circle showing the wheel. If the diameter is 24, what is the radius? If the center is at 18, then how high is the circle above the line for the ground? (Write this into your drawing.) If the car goes around once in 48 seconds and it's been going for 18 seconds, how far around has it gone? If the car (draw a dot) is "directly below" the center of the circle and 6 above the line, then where, relative to the circle, is the dot? What triangle can you form with this information? What is the central angle value? What is the value of the hypotenuse? What then must be the value of the triangle's height? (Add this to the height from the center to get the height above the ground.)