SOLUTION: A hot-air balloon 70 meters above the ground is falling at a constant rate of 6 meters per second while another hot-air balloon 10 meters above the ground is rising at a constant r

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Question 256020: A hot-air balloon 70 meters above the ground is falling at a constant rate of 6 meters per second while another hot-air balloon 10 meters above the ground is rising at a constant rate of 15 meters per second. To the nearest tenth of a second, after how many seconds will the 2 balloons be the same height above the ground?
A. 8.9 B. 6.7 C. 2.9 D. 0.4 E. 0.2

Found 2 solutions by drk, stanbon:
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
We have 2 formulas
(i) 70+-+6S
(ii) 10+%2B+15S
step 1 - set them equal to each other as
(iii) 70+-+6S+=+10+%2B+15S
Solve for S as
60 = 21S
S = 60/21 ~ 2.85 seconds or [C]

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A hot-air balloon 70 meters above the ground is falling at a constant rate of 6 meters per second while another hot-air balloon 10 meters above the ground is rising at a constant rate of 15 meters per second. To the nearest tenth of a second, after how many seconds will the 2 balloons be the same height above the ground?
---
Height of the 1st balloon
h(t) = 70-6t
----
Height of the 2nd balloon
H(t) = 10+15t
--------------------
Equation:
Solve for t:
70-6t=10+15t
21t = 60
t = 60/21 = 2.857 seconds
==============================
Cheers,
Stan H.

A. 8.9 B. 6.7 C. 2.9 D. 0.4 E. 0.2