y: T cosq - mg = may = 0
x: T sinq = mac = mv2/r
[T sinq]/[T cosq] = tanq = [mv2/r]/[mg] = v2/gr.
Then, v = [gr tanq]1/2 = 25.9 m/s
A passenger on a ride at the firemen's field days sits in a chair and is swung around at the end of a cable connected to a central tower. At full speed, the cable makes an angle of 56o with the vertical, and the chair is 46 m from the pole.
1)What is the speed of the passenger?
Please help ASAP!!
1 answer