# A group of students is tracking a friend, Sam, who is riding a Ferris Wheel. They know that Sam reaches a maximum height of 15m and............the minimum height of 2m. The Ferris wheel completes...

A group of students is tracking a friend, Sam, who is riding a Ferris Wheel. They know that Sam reaches a maximum height of 15m and.......

.....the minimum height of 2m. The Ferris wheel completes two full revolutions in 3 minutes. Starting at the lowest point, determine an equation that gives Sam's height 'h' in meters in 't' seconds?

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This can be done using angular motion.

The Ferris wheel starts at the bottom point and complete two revolutions in 3min.

Using angular motion equations

`theta = omega_0t+1/2xxalphaxxt^2`

`omega_0 = 0`

`theta = 2xx2pi = 4pi`

`t = 3min = 180s`

Therefore;

`alpha = (2xx4pi)/(180^2) = 0.00077(rad)/s^2`

It is given that highest and lowest positions located 15m and 2m respectively. So the diameter of the Ferris wheel is (15-2) = 13m

Let us say the wheel is at an angle of theta from the bottom vertical at time t.

At this point;

Height h from bottom point `= 13/2- (13/2)costheta`

`theta = omega_0t+1/2xxalphaxxt^2`

`theta = 1/2xx0.00077xxt^2`

`theta = 0.000388t^2`

Height h from bottom point `= 13/2- (13/2)costheta`

Height h from bottom point `= 13/2- (13/2)cos(0.000388t^2)`

This is the height from the bottom point which is above 2m from ground.

Height from ground `= 2+13/2- (13/2)cos(0.000388t^2)`

*Height from ground `= 8.5-6.5cos(0.000388t^2)` *

*Assumption*

*The Ferris wheel starts from rest at bottom point.*

**Sources:**