The height, h, in metres, above the ground of a rider on a Ferris wheel can be modelled by the equation:

h= 10 sin ((pi/15 t) - 7.5) + 12

where t is the time, in seconds.

At t=0, the rider is at the lowest point. Determine the first two times that the rider is 20 m above the ground, to the nearest hundredth of a second.

1 answer

we want h to be 20
20 = 10 sin ((pi/15 t) - 7.5) + 12
8 = 10 sin ((pi/15 t) - 7.5)
.8 = sin ((pi/15 t) - 7.5)
(pi/15 t) - 7.5) = .927295 or (pi/15 t) - 7.5) = pi - .927295 = 2.214297

Case 1: (pi/15 t) - 7.5) = .927295
pi/15 t = 8.427295
t = 40.237

case 2: (pi/15 t) - 7.5) = 2.214297
t = 46.28235

But the period of your wheel is 2pi/(pi/15) = 30 seconds, so my answers are for the second rotation.

Let’s subtract 30 seconds, to get
times of 10.24 sec and 16.28 seconds

check: if t = 10.24
h = 10sin(15/pi*10.24 - 7.5) + 12
= 20.016 (pretty close)
My other answer also works.