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An object on the top of a hill 100 m high is just visible above the horizon from a station at sea level. The distance between the station and the object is:
- a)38.53 km
- b)3.853 km
- c)3853 km
- d)385.3 km
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
An object on the top of a hill 100 m high is just visible above the ho...
Distance between the station and the object is given by:
Where,
h = height of visible horizon (meters)
D = Distance of visible horizon (km)
D = 38.53 km.
Most Upvoted Answer
An object on the top of a hill 100 m high is just visible above the ho...
Given information:
Height of the hill (h) = 100 m
Distance between the station and the object (d) = ?
Approach:
The problem can be solved using trigonometry. We can use the fact that the line of sight between the station and the object is tangent to the earth's surface at the station. We can form a right-angled triangle with the height of the hill as the opposite side, the distance between the station and the object as the hypotenuse, and the radius of the earth plus the height of the station as the adjacent side.
Solution:
Let R be the radius of the earth and h be the height of the hill. Let x be the distance between the station and the point where the line of sight from the station to the top of the hill is tangent to the earth's surface.
Using Pythagoras theorem, we have:
(R + h)² = x² + d² ...(1)
Also, we have:
tan θ = h/d
θ = tan⁻¹(h/d)
θ = tan⁻¹(h/(√(x² + d²))) ...(2)
From the figure, we have:
tan(π/2 - θ) = R/(R + h)
tan(π/2 - θ) = (R + h - x)/d ...(3)
Using the identity tan(π/2 - θ) = 1/tanθ, we get:
d/tanθ = R + h - x
d/tan(tan⁻¹(h/(√(x² + d²)))) = R + h - x
d = (R + h - x)/tan(tan⁻¹(h/(√(x² + d²))))
d = (R + h - x)/(h/(√(x² + d²)))
d = √(x² + d²)(R + h - x)/h
Solving equation (1) for x, we get:
x = √(R² + 2Rh)
Substituting the value of x in the above equation, we get:
d = √((R² + 2Rh + h²)/h²) × (R + h - √(R² + 2Rh))
d = √(1 + (2Rh + h²)/R²) × (R + h - √(R² + 2Rh))
d = √(1 + (2 × 6.371 × 10⁶ × 100 + 100²)/(6.371 × 10⁶)²) × (6.371 × 10⁶ + 100 - √(6.371 × 10⁶)² + 2 × 6.371 × 10⁶ × 100))
d = 38.53 km (approx)
Therefore, the distance between the station and the object is approximately 38.53 km.
Height of the hill (h) = 100 m
Distance between the station and the object (d) = ?
Approach:
The problem can be solved using trigonometry. We can use the fact that the line of sight between the station and the object is tangent to the earth's surface at the station. We can form a right-angled triangle with the height of the hill as the opposite side, the distance between the station and the object as the hypotenuse, and the radius of the earth plus the height of the station as the adjacent side.
Solution:
Let R be the radius of the earth and h be the height of the hill. Let x be the distance between the station and the point where the line of sight from the station to the top of the hill is tangent to the earth's surface.
Using Pythagoras theorem, we have:
(R + h)² = x² + d² ...(1)
Also, we have:
tan θ = h/d
θ = tan⁻¹(h/d)
θ = tan⁻¹(h/(√(x² + d²))) ...(2)
From the figure, we have:
tan(π/2 - θ) = R/(R + h)
tan(π/2 - θ) = (R + h - x)/d ...(3)
Using the identity tan(π/2 - θ) = 1/tanθ, we get:
d/tanθ = R + h - x
d/tan(tan⁻¹(h/(√(x² + d²)))) = R + h - x
d = (R + h - x)/tan(tan⁻¹(h/(√(x² + d²))))
d = (R + h - x)/(h/(√(x² + d²)))
d = √(x² + d²)(R + h - x)/h
Solving equation (1) for x, we get:
x = √(R² + 2Rh)
Substituting the value of x in the above equation, we get:
d = √((R² + 2Rh + h²)/h²) × (R + h - √(R² + 2Rh))
d = √(1 + (2Rh + h²)/R²) × (R + h - √(R² + 2Rh))
d = √(1 + (2 × 6.371 × 10⁶ × 100 + 100²)/(6.371 × 10⁶)²) × (6.371 × 10⁶ + 100 - √(6.371 × 10⁶)² + 2 × 6.371 × 10⁶ × 100))
d = 38.53 km (approx)
Therefore, the distance between the station and the object is approximately 38.53 km.
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An object on the top of a hill 100 m high is just visible above the horizon from a station at sea level. The distance between the station and the object is:a)38.53 kmb)3.853 kmc)3853 kmd)385.3 kmCorrect answer is option 'A'. Can you explain this answer?
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