A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. In the water what angle does the beam make with the surface

θ₂ = 40.5º

Explanation:

For this exercise we must use the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for the incident ray and index 2 is for the refracted ray

in this case the incident ray has an angle of θ₁ = 60º and the refractive index of the water is

          n₂ = 1,333

        sin θ₂ =

let's calculate

        sin  θ₂ = 1 / 1.3333 sin 60

        sin θ₂ = 0.64968

        θ₂ = sin⁻¹ (0.64968)

        θ₂ = 40.5º

thermal energy is the answer

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft