A laser emits a light beam with a wavelength of 630 nm. The jet passes a liquid with a refractive index of 1.3.
1 answer:
You might be interested in
Answer:
The projectile lands at a distance of 54.2 m from the launching point.
Explanation:
The projectile´s position is described by the following position vector:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upaward direction as positive).
When the projectile lands, its vector position is "r final" (see figure). Notice that the y-component of that vector is 0 m if we place the origin of the frame of reference at the launching point. Then, using the equation for the vertical position of the projectile:
y = y0 + v0 · t · sin α + 1/2 · g · t²
Notice that y0 = 0 because the origin of the frame of reference is located at the launching point. Then, let´s write the equation when y = 0 and solve for the time:
0 = 24,3 m/s · t · sin 32° - 1/2 · 9.8 m/s² · t²
0= t ( 24.3 m/s · sin 32° - 4.9 m/s² · t) Solution 1 t = 0 (the initial position)
0 = 24.3 m/s · sin 32° - 4.9 m/s² · t
- 24.3 m/s · sin 32 / - 4.9 m/s² = t
t = 2.63 s
Now, we can calculate the x- component of the vector r final that is the horizontal distance covered by the projectile:
x = x0 + v0 · t · cos α
x = 0 m + 24.3 m/s · 2.63 s · cos 32°
x = 54.2 m
The projectile lands at a distance of 54.2 m from the launching point.
