Introduction / Context:
This is a classic height and distance problem involving a balloon that moves vertically due to its own ascent and horizontally due to wind. Angles of elevation observed at two different times allow us to compute how far the balloon has moved sideways, which in turn reveals the horizontal wind speed over a particular interval, namely the last five minutes.
Given Data / Assumptions:
- The balloon ascends vertically at 18 km/h.
- Angle of elevation after 10 minutes is 60°.
- Angle of elevation after 15 minutes is 45°.
- The observation point on the ground is fixed.
- We assume the vertical speed is constant, and the horizontal motion during the last five minutes is due to wind at constant speed in that interval.
Concept / Approach:
The key idea is to convert the vertical speed into vertical distances at 10 and 15 minutes, then use trigonometry. At each time, we have a right triangle where the opposite side is the altitude of the balloon and the adjacent side is the horizontal distance from the observer. Using tan of the angle of elevation gives the ratio altitude / horizontal distance. From this we can deduce the horizontal distances at 10 and 15 minutes and then calculate the change in horizontal distance over the last five minutes. Dividing this horizontal displacement by the time interval gives the wind speed.
Step-by-Step Solution:
1. Convert 10 minutes to hours: 10 minutes = 10 / 60 hours = 1 / 6 hour.
2. Vertical speed of balloon = 18 km/h, so altitude after 10 minutes: h1 = 18 * (1 / 6) = 3 km.
3. Similarly, 15 minutes = 15 / 60 hours = 1 / 4 hour.
4. Altitude after 15 minutes: h2 = 18 * (1 / 4) = 4.5 km.
5. Let the horizontal distance from the observer at 10 minutes be x1.
6. For the 10 minute triangle, tan 60° = h1 / x1, so √3 = 3 / x1, hence x1 = 3 / √3 = √3 km.
7. Let the horizontal distance at 15 minutes be x2.
8. For the 15 minute triangle, tan 45° = h2 / x2, so 1 = 4.5 / x2, thus x2 = 4.5 km.
9. Horizontal displacement in the last five minutes = x2 − x1 = 4.5 − √3 km.
10. Numerically, √3 is about 1.732, so displacement ≈ 4.5 − 1.732 = 2.768 km.
11. The time interval is 5 minutes = 5 / 60 hours = 1 / 12 hour.
12. Wind speed = horizontal displacement / time = 2.768 / (1 / 12) ≈ 2.768 * 12 ≈ 33.216 km/h.
13. The closest option to this value is 33 km/h.
Verification / Alternative check:
We can verify correctness by plugging in 33 km/h as the approximate wind speed and checking reasonableness of the distances. In 5 minutes, wind at 33 km/h covers 33 * (5 / 60) ≈ 2.75 km, very close to the more precise 2.768 km we computed. The small difference comes from rounding √3 to 1.732. This confirms that our angle based calculation is consistent with the chosen answer.
Why Other Options Are Wrong:
Option 7 km/h and 11 km/h are far too small, giving horizontal displacements under 1 km in five minutes, which would not reduce the angle of elevation from 60 degrees to 45 degrees. Option 26 km/h is closer but still underestimates the required horizontal shift, which is nearly 2.8 km. Only 33 km/h gives a realistic displacement value that matches the observed change in angle of elevation.
Common Pitfalls:
Learners often forget to convert minutes into hours when using speeds given in km/h, leading to incorrect altitude values. Another common mistake is to interchange opposite and adjacent sides in the tangent ratio, or to confuse tan 60° and tan 45°. Some may incorrectly assume constant horizontal speed over the entire 15 minutes and try to use both angle conditions simultaneously without focusing on the last five minutes, which complicates the solution unnecessarily.
Final Answer:
The approximate speed of the wind during the last five minutes is
33 km/h.
Discussion & Comments