The angles in the equation are not the angles relative to
the horizon but are relative to the "normal" which means that the
line that is perpendicular to the surface.
The angle under the water is 90 – 60 = 30.
n1 for water is 1.33, n2 for air is 1 Which you seem to understand.
(1.33)(sin (30)) = (1.00)(sin (x2))
Rearranging the equation above, will give us: x^2 = sin^-1((1.33
sin 30)/1) = 41.68
But remember that that is the angle relative to the normal
so you have to deduct it from 90 to get the angle relative to the horizon and
you get (90 – 41.68) = 48.32 degrees.
Answer:
Explanation:
The velocity of the swimmer just before touching the water is:
The average force exerted on the diver by the water is determined by the use of the Principle of Energy Conservation and the Work-Energy Theorem:
Answer:
12.5m/s
Explanation:
(Assuming the question was asking for the speed just before it hit the ground)
We can use the first key equation of accelerated motion
Vf^2 = Vi^2+2aΔd
Vf^2 = 0 + 2(9.8)(8) (plugged in values, initial velocity is 0 since the ball was at rest)
Vf^2 = 156.8
Vf = 12.5 (squared both sides)
