Plane Speed Calculation Using Trigonometry

This problem involves calculating the speed of an airplane using angles of elevation and the time taken for a change in angle. We can solve this using basic trigonometry and the formula for speed.

Understanding the Setup

• An observer is standing on open ground.
• A plane is flying horizontally at a constant height.
• The plane is flying away from the observer.
• The observer measures the angle of elevation to the plane at two different times.
• The time elapsed between these measurements is given.

Given Information:

• Constant height of the plane, $h = 100\sqrt{3}$ m.
• Initial angle of elevation, $\alpha = 60^\circ$.
• Final angle of elevation, $\beta = 30^\circ$.
• Time taken for the angle change, $t = 4$ seconds.

Objective:

Calculate the speed of the plane in m/sec.

Applying Trigonometry

Let's visualize the situation. We can form two right-angled triangles:

1. The first triangle is formed by the observer's position (M), the point directly below the plane's initial position (A), and the plane's initial position (P1). The height is P1A, and the horizontal distance is MA ($d_1$). The angle of elevation is $\angle PMA = 60^\circ$.
2. The second triangle is formed by the observer's position (M), the point directly below the plane's final position (B), and the plane's final position (P2). The height is P2B, and the horizontal distance is MB ($d_2$). The angle of elevation is $\angle PMB = 30^\circ$.

Since the plane is flying away from the observer, the distance MB ($d_2$) will be greater than MA ($d_1$). The distance the plane traveled horizontally is $AB = d_2 - d_1$. The speed ($v$) is this distance divided by the time ($t$).

Step 1: Calculate the initial horizontal distance ($d_1$)

Using the first triangle ($\triangle PMA$):

We know that $\tan(\text{angle}) = \frac{\text{Opposite}}{\text{Adjacent}}$.

$\tan(\alpha) = \frac{h}{d_1}$

$\tan(60^\circ) = \frac{100\sqrt{3}}{d_1}$

We know $\tan(60^\circ) = \sqrt{3}$.

So, $\sqrt{3} = \frac{100\sqrt{3}}{d_1}$.

Solving for $d_1$: $d_1 = \frac{100\sqrt{3}}{\sqrt{3}} = 100$ m.

Step 2: Calculate the final horizontal distance ($d_2$)

Using the second triangle ($\triangle PMB$):

$\tan(\beta) = \frac{h}{d_2}$

$\tan(30^\circ) = \frac{100\sqrt{3}}{d_2}$

We know $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.

So, $\frac{1}{\sqrt{3}} = \frac{100\sqrt{3}}{d_2}$.

Solving for $d_2$: $d_2 = 100\sqrt{3} \times \sqrt{3} = 100 \times 3 = 300$ m.

Step 3: Calculate the distance traveled by the plane

The plane moved horizontally from a distance $d_1$ to $d_2$. The distance traveled is the difference between these two distances.

Distance traveled $= d_2 - d_1 = 300 \text{ m} - 100 \text{ m} = 200$ m.

Step 4: Calculate the speed of the plane

Speed is defined as distance traveled per unit time.

Speed, $v = \frac{\text{Distance traveled}}{\text{Time taken}}$

$v = \frac{200 \text{ m}}{4 \text{ s}}$

$v = 50$ m/sec.

Conclusion

The speed of the plane is 50 m/sec.