distance of horizon from me
how to calculate the distance of horizon(a place where land and sky meet) from me.my height is 1.5m
Note by
Sri Venkata Vivek Dhulipala
8 years, 3 months ago
No vote yet
3 votes
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
I think this problem can be approached as follows (assuming that the Earth is a sphere).
One can see an object with negligible height on Earth if and only if it lies closer to the observer than the horizon. So we can define horizon as the limit beyond which no object (of negligible height) located on the surface of the Earth can be seen. Now let the radius of the Earth be r, and the height of the observer be h . We have to find the maximum distance from the observer where the observer can see an object of negligible height if placed. The point where the object will be located has to be the point where the tangent drawn from the head of the observer touches the Earth. This is because if the object is placed at a point further from the point where the tangent touches the Earth, the straight line joining the head of the observer to that point will touch the Earth at another point which is closer to the observer than the original point. So light rays will be blocked and the object cannot be seen. Thus we can conclude that the point beyond which no object of negligible height can be seen is the point where the tangent drawn from the head of the observer touches the Earth. Simple application of Pythagoras Theorem gives this distance to be sqrt{(r+h)^2 - r^2}= sqrt{r(2r+h)}. This is the straight line distance from the observer to the horizon.
If the distance along the surface of the Earth has to be found out, at first the angle subtended by this straight line of length sqrt{(r+h)^2 - r^2}= sqrt{r(2r+h)} has to be found out. Simple application of cosine rule gives this angle to be cos^(-1) {3r^2 - (r+h)^2}/(2r^2). Assuming that this angle is measured in radians, we can find out the arc distance using the definition of radian. Simple argument gives the arc length to be r cos^(-1)[{3r^2 - (r+h)^2}/(2r^2)].
Putting r= 6370 m, h= 1.5 m, it can be seen that the straight line distance to the horizon is ~~ 138.246 m, while the distance along the surface of the Earth is ~~ 138.250 m. Since the height 1.5 m is so negligible compared to the radius of the Earth, the straight line distance and the distance along the surface of the Earth are almost equal.
Log in to reply
your explanation is right.But your calculation is wrong answer comes out to be 5000m approx.It is a previous INAO question
Log in to reply
Sorry, I actually made a mistake. The radius of the Earth is 6370 km, not 6370 m. Now substituting yields a value close to 5 km. A similar problem appeared in a Brilliant weekly set of mechanics. It is here (https://brilliant.org/i/MT9Nig/).