The launch site for trigon balloon co. is 250 ft above sea level. A hot air balloon is launched from the site and begins to rise at a rate of 110 ft/min. At the same time, another balloon 2200 ft above sea level begins to descend at a rate of 150 ft/min.
A. How long will it be unit the balloons are at the same elevation?
B. What will their elevation be then?
I don't under stand how to do 1/3 + 1/10 = what!
To find the time until the balloons are at the same elevation, we can set up an equation based on their rates of ascent and descent.
Let's define:
- "x" as the time (in minutes) since the balloons started moving
- "h₁" as the height (in feet) of the balloon being launched from trigon balloon co.'s launch site
- "h₂" as the height (in feet) of the descending balloon
Based on the given information, we have the following equations:
- h₁ = 250 + 110x (the height of the ascending balloon)
- h₂ = 2200 - 150x (the height of the descending balloon)
A. To find the time until the balloons are at the same elevation, we need to solve the equation h₁ = h₂. Substituting the given expressions for h₁ and h₂, we get:
250 + 110x = 2200 - 150x
Simplifying, we have:
260x = 1950
Dividing both sides by 260, we find:
x ≈ 7.50
So, it will be approximately 7.50 minutes until the balloons are at the same elevation.
B. To find their elevation at that time, substitute the value of x back into either h₁ or h₂ equation. Taking h₁ as an example:
h₁ = 250 + 110x
h₁ ≈ 250 + 110(7.50)
h₁ ≈ 250 + 825
h₁ ≈ 1075
Therefore, at that time, the elevation of both balloons will be approximately 1075 feet above sea level.
balloon#1 = 250 + 110t
balloon#2 = 2200 - 150t
When they reach the same height,
250 + 110t = 2200 - 150t
260t = 1950
t = 7.5 ----- 7 min, 30 seconds
how high? plug t = 7.5 into either equation, you must get the same result for each case