Question

Bizarro, located at Six Flags New England, in Agawam, MA, is the name of one of the top-rated roller coasters in the country. The first drop on the roller coaster brings you down 67 meters. In this problem, use g = 9.80 m/s2.
(a) Assuming you are at rest at the top of the hill before the coaster drops precipitously down the incline (the maximum angle with respect to the horizontal is 72°!), and assuming you can neglect air resistance, how fast are you going at the bottom of the hill? Express your answer in miles per hour.

Answer 1

(b) There are some resistive forces acting as the roller coaster comes down, of course, so let's say you actually reach the bottom of the hill with a speed of merely 76.0 miles per hour. In addition, let's say that the bottom of the hill is a vertical circular arc with a radius of 49.0 m. Figure out the size of the upward normal force acting on you at the bottom of the hill, and then divide by the force of gravity acting on you—this will give you the "g-force" (in units of g, the acceleration due to gravity), you experience at the bottom of the hill. (Just put in a number—don't include the g.)

(c) From the bottom of the first hill, the roller coaster goes up a second hill. Let's say that the peak of the second hill is 48.0 m higher than the bottom of the first hill, and that the shape of the second hill is that of a vertical circular arc with radius of 34.0 m. If resistive forces acting on the coaster cause a loss of 5% of the coaster's mechanical energy as it moves from the bottom of the first hill to the top of the second hill, what is your speed as you pass over the top of the second hill? Note that, this time, we're looking for the speed in meters per second.

(d) If your mass happens to be 50.0 kg, what is the magnitude of the normal force acting on you at the top of the second hill? Note that the normal force you're used to feeling is 490 N.

Answer 2

where you stuck at? ^_^

Answer 3

Parts B and C

Answer 4

k, it took me a second but I think I get part B

you'll need to use the equation
\[\frac{ VelocityTangent^2 }{ Radius } = Acceleration\]
does that help, or should I walk through it more wiht you?

Answer 5

Yeah if you explain it more that would really help, I dont' really understand this problem to be honest.