Question

Bill and Jerry are both taking their hot-air balloons to the balloon show.
They are good friends and like to share experiences. The especially like that when
they get to fly their balloons next to each other. Unfortunately they did not get
cleared to depart at the same time. Bill got to go up first and had already started
to descend by the time that Jerry was cleared. Five seconds before Jerry takes
off Bill’s balloon basket was 1032 feet above the ground falling at a constant rate.
Five seconds after Jerry takes off Bill’s basket was 985 feet above the ground
and Jerry’s balloon basket was 57 feet off the ground and rising at the same
constant rate the whole time. To the nearest tenth of a second, how long after
Jerry’s balloon took off will the height of the two balloon baskets be the same
height?

Answer 1

9514 1404 393

Answer:

  62.6 seconds

Step-by-step explanation:

We can use the 2-point form of the equation for a line to write equations for the heights of the balloons.

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

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For Bill's balloon, the two given points are ...

  (t, h) = (-5, 1032) and (5, 985)

Then the equation is ...

  h = (985 -1032)/(5 -(-5))(t -(-5)) +1032

  h = -47/10(t +5) +1032

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For Jerry's balloon, the two given points are ...

  (t, h) = (0, 0) and (5, 57)

Then the equation is ...

  h = (57-0)/(5-0)(t -0) +0

  h = (57/5)t

__

The heights are equal when ...

  -47/10(t +5) +1032 = (57/5)t

  1008.5 = t(57/5 +47/10) = 16.1t

Then the time until heights are the same is ...

  t = 1008.5/16.1 ≈ 62.6 . . . seconds