The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 20 seco

Q: Wait, why do we need to convert the speed to m/s? Can't we just work in km/h?

That is a great question! In physics and mathematics, consistency is the bedrock of accuracy. Since the time given is in seconds, working with km/h would lead to a mismatch in units. If you kept the speed in km/h, your distance would be in kilometers, but your height would likely be calculated in meters. Always convert to SI units (meters and seconds) at the very beginning to avoid these hidden traps.

Q: Why are we using the tangent function instead of sine or cosine?

Excellent observation. We use the tangent function because it directly relates the two sides we are interested in: the perpendicular (the height $h$) and the base (the horizontal distance). Sine and cosine involve the hypotenuse (the line of sight), which is an unknown variable we don't need to find. By choosing tangent, we eliminate the hypotenuse entirely, making our algebra much cleaner!

Q: Why is the base of the second triangle $x + 2400$ and not just $2400$?

Think about the geometry. The angle of elevation is measured from the *same* point A on the ground. The jet moves from point C to point D. The distance $2400 \text{ m}$ is only the distance between the two positions of the jet. The base of the second triangle must be the *total* horizontal distance from the observer to the point directly below the jet's second position, which includes the initial distance $x$ plus the distance traveled $2400 \text{ m}$.

Question

The angle of elevation of a jet plane from a point A on the ground is $60^{\circ}$. After a flight of 20 seconds at speed of 432 km/ hour, the angle of elevation changes to $30^{\circ}$. If the jet plane is flying at a constant height, then its height is:

Options:

(A) $1200\sqrt{3}$ m

(B) $1800\sqrt{3}$ m

(C) $3600\sqrt{3}$ m

(D) $2400\sqrt{3}$ m

Sigma Percentile · Accepted Answer

### The Geometry of Flight: A Journey Through the Skies

Imagine you are standing on a vast, open field, looking up at the sky. A jet plane streaks across the horizon, a silver needle threading through the blue.

You are not just an observer; you are a navigator of physics. Today, we are going to solve a classic JEE problem that turns this simple act of watching a plane into a beautiful exercise in trigonometry and algebraic elegance.

### Phase 1: The Physics of Motion

Before we even draw a single triangle, we must respect the physics of the situation. The problem tells us the jet is flying at a speed of $432 	ext{ km/h}$.

But look closely at the time: $20 	ext{ seconds}$. We have a mismatch! If we try to mix kilometers per hour with seconds, our answer will be as lost as a plane without a flight plan.

We must convert the speed into meters per second. We use our conversion factor:

$$v = 432 	imes \frac{5}{18} 	ext{ m/s}$$

Calculating this, $18$ goes into $432$ exactly $24$ times, and $24 	imes 5$ gives us a crisp $120 	ext{ m/s}$.

Now, with the speed in meters per second, we can find the distance $d$ the jet travels in $20$ seconds:

$$d = 120 	imes 20 = 2400 	ext{ m}$$

This is the horizontal distance between the two points where we observe the jet.

### Phase 2: The Geometry of Sight

Now, let us visualize the scene. We have two right-angled triangles sharing the same height $h$.

Let the initial horizontal distance from you to the point directly below the jet be $x$. In our first triangle, the angle of elevation is $60^{\circ}$.

Using the tangent function, we have:

$$	an 60^{\circ} = \frac{h}{x}$$

Since $	an 60^{\circ} = \sqrt{3}$, we find that $x = \frac{h}{\sqrt{3}}$. This is our first key, our Equation 1.

Next, the jet flies forward. The new angle of elevation is $30^{\circ}$. The height $h$ remains constant, but the base of our triangle has grown.

It is now the original distance $x$ plus the distance the jet traveled, $2400 	ext{ m}$. So, for our second triangle, we have:

$$	an 30^{\circ} = \frac{h}{x + 2400}$$

Knowing that $	an 30^{\circ} = \frac{1}{\sqrt{3}}$, we get:

$$\frac{1}{\sqrt{3}} = \frac{h}{x + 2400}$$

This simplifies to $x + 2400 = h\sqrt{3}$. This is our Equation 2.

### Phase 3: The Algebraic Dance

We have two equations and two unknowns. It is time to solve.

We substitute $x = \frac{h}{\sqrt{3}}$ from Equation 1 into Equation 2:

$$\frac{h}{\sqrt{3}} + 2400 = h\sqrt{3}$$

Now, let us isolate $h$. We move the $h$ terms to one side:

$$2400 = h\sqrt{3} - \frac{h}{\sqrt{3}}$$

To subtract these, we find a common denominator of $\sqrt{3}$:

$$2400 = \frac{3h - h}{\sqrt{3}} \Rightarrow 2400 = \frac{2h}{\sqrt{3}}$$

Finally, we solve for $h$:

$$2h = 2400\sqrt{3} \Rightarrow h = 1200\sqrt{3} 	ext{ m}$$

### The Takeaway

Look at that result: **$1200\sqrt{3} 	ext{ m}$**. It is not just a number; it is the height of the jet, derived purely from the relationship between angles and distances.

The beauty of this problem lies in how it forces us to bridge the gap between physical motion and geometric abstraction. Whenever you face a problem like this, remember: visualize the triangles, respect the units, and trust the algebra.

You are not just solving for $h$; you are mastering the language of the universe.

Select Answer:

Visualized Solution

Initial Position of the Jet

Final Position of the Jet

Converting Speed to Standard Units

Calculating Speed in m/s

Calculating Horizontal Distance

Defining the Base of the First Triangle

Analyzing the First Triangle

Equation from the First Triangle

Analyzing the Second Triangle

Equation from the Second Triangle

Substituting $x$

Isolating $h$

Simplifying the Expression

Final Calculation

The Sigma Insight: Heights and Distances

The Geometry of Flight

Phase 1

Phase 2

Phase 3

The Takeaway

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