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2 answers:
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Answer:
Explanation:
Data provided in the question:
Height above the ground, H= 5.0m
Range of the ball, R= 20 m
Initial horizontal velocity =
Initial vertical velocity= (Since ball was thrown horizontally only)
Acceleration acting horizontally, = 0 m/s² [ Since no acceleration acts horizontally) ]
Vertical Acceleration, = 9.8 m/s² (Since only gravity acts on it)
Let 't' be the time taken to reach ground
Therefore, using equations of motion, we have
Then using Equations of motion for horizontal motion,
Answer
Given,
refractive index of film, n = 1.6
refractive index of air, n' = 1
angle of incidence, i = 35°
angle of refraction, r = ?
Using Snell's law
n' sin i = n sin r
1 x sin 35° = 1.6 x sin r
r = 21°
Angle of refraction is equal to 21°.
Now,
distance at which refractive angle comes out
d = 2.5 mm
α be the angle with horizontal surface and incident ray.
α = 90°-21° = 69°
t be the thickness of the film.
So,
t = 2.26 mm
Hence, the thickness of the film is equal to 2.26 mm.