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This reduces the shadow zone, but causes errors in distance and height measuring. In practice, to find , one must be using a value of 8.5·10 3 km for the effective Earth's radius (4/3 of it), instead of the real one. [2] So the equation becomes:
The height of the elevated point plus the Earth radius form its hypotenuse. If both the eyes and the object are raised above the reference plane, there are two right-angled triangles. If both the eyes and the object are raised above the reference plane, there are two right-angled triangles.
For example, for an observer B with a height of h B =1.70 m standing on the ground, the horizon is D B =4.65 km away. For a tower with a height of h L =100 m, the horizon distance is D L =35.7 km. Thus an observer on a beach can see the top of the tower as long as it is not more than D BL =40.35 km away.
If the height h is given in feet, and the distance d in statute miles, d ≈ 1.23 ⋅ h {\displaystyle d\approx 1.23\cdot {\sqrt {h}}} R is the radius of the Earth, h is the height of the ground station, H is the height of the air station d is the line of sight distance
Also, the velocities in the directions perpendicular to the frame changes are affected, as shown above. This is due to time dilation, as encapsulated in the dt/dt′ transformation. The V′ y and V′ z equations were both derived by dividing the appropriate space differential (e.g. dy′ or dz′) by the time differential.
We are interested in the time when the projectile returns to the same height it originated. Let t g be any time when the height of the projectile is equal to its initial value. 0 = v t sin θ − 1 2 g t 2 {\displaystyle 0=vt\sin \theta -{\frac {1}{2}}gt^{2}}
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