Oscar is also watching an airplane, P, fly directly over his head - Leaving Cert Mathematics - Question c - 2022

In Diagram 2, the sound takes 41 seconds to reach Oscar after the airplane has flown a certain distance at a speed of 255 meters per second. The distance it has traveled is:

$ext{Distance} = speed \times time = 255 m/s \times 41 s = 10,455 m$

The triangle formed shows that we can use the tangent function:

$\tan(\theta) = \frac{10}{10,455}$

Calculating the angle:

$\theta = \tan^{-1}\left(\frac{10}{10,455}\right)$

This gives us approximately:

The equation relates the positions of the airplane and the distances covered by sound and motion. The left side, √100 + a², represents the distance traveled by sound from P3 to P4, where 100 is the vertical distance and a is horizontal. On the right, 2d represents the the total distance covered horizontally when factoring in the airplane's speed and time based on the known sound speed.

To solve the equation:

$\sqrt{100 + a^2} = 2d$

We first need to square both sides:

$100 + a^2 = 4d^2$

Assuming we know the vertical position and utilizing the speed of sound, we can simplify it.

After simplifying, we find:

$d = \sqrt{\frac{(100 + a^2)}{4}}$

Calculating arrives at an approximate value of 4.0 km (1 D.P.).