Aroller coaster starting from rest descends 35 meters in its initial drop and then rises 23 meters when it goes over the first hill, which has a circular shape over the top. if a passenger at the top of the hill feels an apparent weight equal to one-half of her normal weight, what is the radius of curvature of the first hill? neglect any frictional losses.

Radius of hill = r = 48 m

Explanation:

Given data:

Initial height = = 35 m

Final height =   = 23m

Radius = r = ?

We know that

Change in potential energy = ΔP.E = mg(Δh)

ΔP.E = mg( - )

ΔP.E = mg(35 - 23)

ΔP.E = 12 mg

We know that

Change in Kinetic Energy = ΔK.E = 0.5 mv²

According to energy of conservation

ΔK.E = ΔP.E

0.5 mv² = 12 mg

v² = 24 g  Eq 1

As she feels an apparent weight which is one-half of her normal weight

So,

Apparent weight = 0.5 normal weight

mg - mv²/r = 0.5 mg

0.5 g = v²/r

v² = 0.5 gr   Eq 2

Compare Eq 1 and Eq 2

0.5 gr = 24 g

r = 48 m

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer