An aeroplane flying at a constant speed, parallel to the horizontal ground, √3 km above it, is obser

Q: Wait, why do we assume the plane is moving in a straight line parallel to the ground?

The problem explicitly states the aeroplane is flying at a 'constant speed, parallel to the horizontal ground.' In physics problems, this is our cue to simplify the motion into a 1D horizontal displacement problem. If it weren't parallel, we would need to account for changing altitude, which would make the trigonometry much more complex!

Q: Why can't I just use the average angle of elevation?

Trigonometric functions like $\tan(\theta)$ are non-linear. Averaging the angles would lead to a completely incorrect horizontal distance. You must calculate the distance for each angle independently and then find the difference between those distances.

Q: I keep getting the wrong speed. Is it the unit conversion?

This is the most common trap! The distance is in kilometers and the time is in seconds. You must convert the time into hours by dividing by $3600$ (since $1$ hour $= 3600$ seconds) to get the speed in km/hr. Always check your units before the final division.

Q: Does the position of the observer matter relative to the path of the plane?

As long as the observer is on the ground and the plane's path is a straight line, the geometry holds. We are essentially looking at a 2D cross-section of the 3D space. The perpendicular distance from the observer to the plane's flight path remains constant.

Q: Why is the distance $d = x_2 - x_1$?

Imagine the plane moving from point A to point B. The horizontal distance from the observer to the plane's projection on the ground changes from $x_1$ to $x_2$. Since the plane is moving away from the observer, the horizontal distance increases, so the difference $x_2 - x_1$ gives the exact horizontal displacement covered during that time interval.

Question

An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt{3}$ km above it, is observed at an elevation of $60^\circ$ from a point on the ground. If, after five seconds, its elevation from the same point, is $30^\circ$, then the speed (in km/hr) of the aeroplane, is :

Options:

(A) 750

(B) 1440

(C) 1500

(D) 720

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in an open field, looking up at the vast, clear sky. You spot an aeroplane, a tiny silver speck moving with grace and precision. It is flying at a constant altitude of $\sqrt{3}$ km, cruising parallel to the ground.

This is a classic problem of kinematics and trigonometry that tests your ability to translate a physical scenario into a precise mathematical model. Let’s break this down together.

### The First Snapshot

At the very first moment of observation, let’s call this $t = 0$, the plane is at position $A$. You look up at an angle of $60^{\circ}$.

To solve this, we drop a perpendicular from the plane to the ground, creating a right-angled triangle. Let the horizontal distance from your feet to the point directly beneath the plane be $x_1$. Using the definition of tangent, we have:

$$
	an(60^{\circ}) = \frac{	ext{Altitude}}{	ext{Horizontal Distance}} = \frac{\sqrt{3}}{x_1}
$$

Since we know $	an(60^{\circ}) = \sqrt{3}$, the equation simplifies beautifully: $\sqrt{3} = \frac{\sqrt{3}}{x_1}$. This tells us that $x_1 = 1$ km. The plane is currently $1$ km away from you in the horizontal plane.

### The Second Snapshot

Five seconds pass. The plane has moved forward to a new position, $B$. Now, when you look up, the angle of elevation has dropped to $30^{\circ}$.

The plane is still at the same altitude of $\sqrt{3}$ km, but it is further away. Let the new horizontal distance be $x_2$. Again, we apply our trigonometric toolkit:

$$
	an(30^{\circ}) = \frac{\sqrt{3}}{x_2}
$$

Knowing that $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$, we substitute this in: $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{x_2}$. Solving for $x_2$, we find $x_2 = 3$ km. The plane has traveled from a horizontal position of $1$ km to $3$ km.

### The Velocity Calculation

Now, we reach the heart of the problem. The distance $d$ covered by the plane in those $5$ seconds is the difference between these two horizontal positions:

$$
d = x_2 - x_1 = 3 - 1 = 2 	ext{ km}
$$

We have the distance, and we have the time. To find the speed in km/hr, we must convert our time $t = 5$ seconds into hours:

$$
t = \frac{5}{3600} 	ext{ hours}
$$

Finally, we use the fundamental definition of speed, $v = \frac{d}{t}$:

$$
v = \frac{2}{\frac{5}{3600}} = \frac{2 	imes 3600}{5}
$$

Calculating this, we get $v = 2 	imes 720 = 1440$ km/hr. The final speed of the aeroplane is **$1440$ km/hr**.

### The Takeaway

Look at how the complexity melted away. By breaking the motion into two distinct snapshots and using the elegance of trigonometry, we turned a dynamic flight path into a static geometric problem.

Never fear the complexity of a problem; instead, look for the triangles hidden within the motion. You have the tools, you have the logic, and now you have the experience. Keep practicing, and soon, these problems will feel like second nature.

Select Answer:

Visualized Solution

Visualizing the Scenario

First Observation Point $A$

Setting up $△ P A A^{'}$

Calculating Distance $x_{1}$

Second Observation Point $B$

Setting up $△ P B B^{'}$

Calculating Distance $x_{2}$

Distance Traveled $d$

Calculating $d$

Speed Formula Setup

Unit Conversion for Time

Substituting Values

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The First Snapshot

The Second Snapshot

The Velocity Calculation

The Takeaway

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