SOLUTION: Hello, my name is niya. I am in the eighth grade and taking algebra honors. I need help with this word problem. A hot air ballon at 300 feet begins to rise at the rate of 100 feet

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Question 487145: Hello, my name is niya. I am in the eighth grade and taking algebra honors. I need help with this word problem. A hot air ballon at 300 feet begins to rise at the rate of 100 feet per minute. At the same time, a second hot-air ballon at 2,00 feet starts to descend at the rate of 150 feet per minute.
a. When will the ballons be at the same height?
b. What is the height?
c. What is the height of the ascending ballon when the descending ballon reaches the grouond?
Found 2 solutions by Edwin McCravy, josmiceli:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Something is wrong with your problem, Niya. You have the rising balloon
starting at 300 ft which is higher than the descending balloon at 200 feet, so
they will never be at the same height. I notice you typed "2,00" which I took
to be 200. Please correct the error and we will help you.


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In this problem, the important thing is the
difference in their heights and the fact that
they are approaching each other at different
rates.
Each balloon has it's own equation
For rising balloon,
(1) +d%5B1%5D+=+r%5B1%5D%2At+
For descending balloon,
(2) +d%5B2%5D+=+r%5B2%5D%2At+
Note that t in each equation is the time it
takes for them to be at the same height.
Initially, they are +2000+-+300+=+1700+ ft apart
so,
(3) +d%5B1%5D+%2B+d%5B2%5D+=+1700+
+r%5B1%5D+=+100+ ft/min
+r%5B2%5D+=+150+ ft/min
-----------------
(1) +d%5B1%5D+=+100t+
(2) +d%5B2%5D+=+150t+
(3) +d%5B1%5D+%2B+d%5B2%5D+=+1700+
I can say
+100t+%2B+150t+=+1700+
+250t+=+1700+
+t+=+1700%2F250+
+t+=+6.8+ min
and
(1) +d%5B1%5D+=+100t+
(1) +d%5B1%5D+=+100%2A6.8+
(1) +d%5B1%5D+=+680+
and
(2) +d%5B2%5D+=+150t+
(2) +d%5B2%5D+=+150%2A6.8+
(2) +d%5B2%5D+=+1020+
The rising balloon starts at +300+ ft and
rises +680+ ft to a height of
+300+%2B+680+=+980+ ft
The descending balloon falls +1020+ ft
to reach a height of +2000+-+1020+=+980+ ft
(a) They will be at the same height in 6.8 mi
(b) The height will be 980 ft
-----------------------
The descending balloon needs to drop 2,000 ft
to reach the ground, so
(2) +d%5B2%5D+=+150t+
(2) +2000+=+150t+
(2) +t+=+13.333+ min
Now find out how hight the rising balloon is in this time
(1) +d%5B1%5D+=+100t+
(1) +d%5B1%5D+=+100%2A13.333+
(1) +d%5B1%5D+=+1333.333+
This balloon started at +300+ ft, so
(c)
+300+%2B+1333.333+=+1633.333+ ft is the
height of the rising balloon
------------------------
Note that it would have been more correct to make
rising displacements and velocities positive
and falling ones negative. I avoided this, however.
Hope the answers are right.