Light enters water from air at an angle of 25° with the normal, θ1. if water has an index of refraction of 1.33, determine θ2.
a) 18.5°
b) 34.2°
c) 31.8°
d) 71.5°

By Snell's law:

η = sini / sinr.        i = 25,  η = 1.33

1.33 = sin25° / sinr

sinr = sin25° / 1.33 = 0.4226/1.33 = 0.3177    Use a calculator.

r = sin⁻¹(0.3177)

r ≈ 18.52°

Option A.

God's grace.

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward