A balloon is rising vertically above a leve straight road at a constant rateof 1 ft/sec. Just when the balloon is 65 feet above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it.

What is the distance between the balloon and the bicyclist three seconds later?

What is the distance between the balloon and the bicyclist after another 3 seconds have past?

2. Hello, JaredMims!

A balloon is rising vertically above a level road at a constant rate of 1 ft/sec.
Just when the balloon is 65 feet above the ground,
a bicycle moving at a constant rate of 17 ft/sec passes under it.

(a) What is the distance between the balloon and the bicyclist three seconds later?
Code:
    A *
|\
t | \
|  \
C *   \
|    \ d
|     \
65 |      \
|       \
|        \
D * - - - - * B
17t
The balloon is at $C$, rising at 1 ft/sec.
In the next $t$ seconds, it rises $t$ feet to $A.$
. . $AD \,=\,t + 65$ feet.

The cyclist is at $D$, moving at 17 ft/sec.
In the next $t$ seconds, it moves $17t$ feet to $B.$
. . $DB \,=\,17t$

In 3 seconds, the balloon is 68 feet high, the cyclist moved 51 feet.

Pythagorus give us the distance: . $d \:=\:\sqrt{68^2+51^2} \:=\:\sqrt{7225} \:=\:85\text{ feet.}$

(b) What is the distance between the balloon and the bicyclist after another 3 seconds have passed?

After 6 seconds, the balloon is 71 feet high, the cyclist moved 102 feet.

Pythagorus says: . $d \:=\:\sqrt{71^2 + 102^2} \:\approx\:126\text{ feet}$

3. Originally Posted by JaredMims
A balloon is rising vertically above a leve straight road at a constant rateof 1 ft/sec. Just when the balloon is 65 feet above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it.

What is the distance between the balloon and the bicyclist three seconds later?

What is the distance between the balloon and the bicyclist after another 3 seconds have past?
this is all Pythagoras ... you just have to figure out the height of the balloon and where the bicycle is relative to the point directly beneath the balloon at the specified times.