Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s
There are two conditions necessary for total internal reflection, which is when light hits the boundary between two mediums and reflects back into its original medium:
Light is about to pass from a more optically dense medium (slower) to a less optically dense medium (faster).
The angle of incidence is greater than the defined critical angle for the two mediums, which is given by:
θ = sin⁻¹(/
)
Where θ = critical angle, = refractive index of faster medium,
= refractive index of slower medium.
Choice C gives one of the above necessary conditions.