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Answer :
The hoop will go up the incline 1.25 m and will be on the incline for 0.67 s before arriving back at the bottom.
Find the velocity of the hoop as it starts up the incline using conservation of energy:
(1/2)mv^2 = (1/2)Iω^2 + mgh
where m is the mass of the hoop, v is its initial velocity, I is the moment of inertia, ω is the angular velocity, and h is the height it will reach. Since it's a hoop, I = mr^2 where r is the radius. Solving for ω and using v = rω, we get ω = v/r and
h = (v^2/2g)(1 - cos θ) = 1.25 m.
Find the time it takes to travel up and back down the incline:
t = 2(vsin θ)/(gcos θ) = 0.67 s.
So hoop will go up the incline 1.25 m and will be on the incline for 0.67 s before arriving back at the bottom.
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