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Part of a roller-coaster ride involves coasting down an incline and entering a loop 30.0 m in diameter. For safety considerations, the roller coasters speed at the top of the loop must be such that the force of the seat on a rider is equal in magnitude to the riderâ€™s weight. From what height above the bottom of the loop must the roller coaster descend to satisfy this requirement?

R = 15 meters

Ac = w^2 R = v^2/R

Ac = 2 g to balance weight plus to give additional mg force

9.8 * 2 = v^2/15
v^2 = 294 m^2/s^2
v at top of loop = 17.1 m/s

potential energy at top of loop = m g h
= m (9.8)(30) = 294 m
kinetic energy at top of loop = (1/2) m (294) = 147 m
total energy at top of loop = 441 m

we need 441 m Joules of potential energy at start to make it
m g h = 441 m
9.8 h = 441
h = 45 meters high

Thank you!

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