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September 30, 2014

September 30, 2014

Posted by **-** on Monday, November 16, 2009 at 7:54pm.

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.

- Math -
**Reiny**, Monday, November 16, 2009 at 9:40pmwe 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.

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