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This is a related rates problem involving the balloon's vertical movement and the bicycle's horizontal movement. Let's break it down step by step:

### Given:
- The balloon is rising vertically at a rate of \( \frac{dy}{dt} = 5 \) ft/sec.
- The bicycle is moving horizontally at a rate of \( \frac{dx}{dt} = 16 \) ft/sec.
- At the moment the bicycle is directly under the balloon, the balloon is 98 ft above the ground.
- We need to find the rate at which the distance between the balloon and the bicycle is increasing after 6 seconds.

### Step 1: Define Variables
- Let \( y(t) \) be the height of the balloon after \( t \) seconds. Initially, \( y(0) = 98 \) feet, and the balloon is rising at 5 ft/sec. So, after \( t \) seconds, the height of the balloon is:
  \[
  y(t) = 98 + 5t
  \]
- Let \( x(t) \) be the horizontal distance traveled by the bicycle after \( t \) seconds. Since the bicycle moves at 16 ft/sec, the horizontal distance after \( t \) seconds is:
  \[
  x(t) = 16t
  \]
- Let \( s(t) \) be the distance between the balloon and the bicycle at any time \( t \). By the Pythagorean theorem:
  \[
  s(t) = \sqrt{x(t)^2 + y(t)^2}
  \]

### Step 2: Find the rate of change of \( s(t) \)
We need to find \( \frac{ds}{dt} \) at \( t = 6 \) seconds.

From the Pythagorean theorem:
\[
s^2 = x^2 + y^2
\]
Differentiate both sides with respect to \( t \):
\[
2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Simplifying:
\[
s \frac{ds}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]
\[
\frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s}
\]

### Step 3: Evaluate at \( t = 6 \)
- At \( t = 6 \) seconds, the horizontal distance traveled by the bicycle is:
  \[
  x(6) = 16 \times 6 = 96 \text{ ft}
  \]
- The height of the balloon at \( t = 6 \) seconds is:
  \[
  y(6) = 98 + 5 \times 6 = 128 \text{ ft}
  \]
- The distance between the balloon and the bicycle at \( t = 6 \) seconds is:
  \[
  s(6) = \sqrt{x(6)^2 + y(6)^2} = \sqrt{96^2 + 128^2} = \sqrt{9216 + 16384} = \sqrt{25600} = 160 \text{ ft}
  \]

Now substitute these values into the expression for \( \frac{ds}{dt} \):
\[
\frac{ds}{dt} = \frac{96 \times 16 + 128 \times 5}{160}
\]
\[
\frac{ds}{dt} = \frac{1536 + 640}{160} = \frac{2176}{160} = 13.6 \text{ ft/sec}
\]

### Final Answer:
The distance between the balloon and the bicycle is increasing at a rate of **13.6 ft/sec** after 6 seconds.

Would you like further details or have any questions?  
Here are 5 related questions to expand your understanding:
1. How would the rate change if the bicycle were moving faster or slower?
2. What if the balloon were rising at a variable rate instead of a constant rate?
3. How can this problem be applied to real-life scenarios like drones and moving vehicles?
4. Can we use this method to calculate the distance at any arbitrary time \( t \)?
5. What happens to the distance between the bicycle and the balloon if both stop moving?

**Tip:** Always check that you're differentiating both sides of an equation correctly, especially in related rates problems.

A balloon is rising vertically above a level, straight road at a constant rate of 5 ft/sec. Just when the balloon is 98 ft above the ground, a bicycle moving at a constant rate of 16 ft/sec passes under it. How fast is the distance between the bicycle and balloon increasing 6 seconds later?

Solve a related rates problem where a balloon rises vertically at 5 ft/sec and a bicycle moves horizontally at 16 ft/sec. Learn how to calculate the rate of change in the distance between them after 6 seconds using the Pythagorean theorem and differentiation.

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