A child releases a balloon from a height of 6ft and it ascends at a rate of 5ft per second. A second balloon starts at a height of 51 ft and descends at a rate of 2 ft per second. After how many seconds will both balloons be at the same height above the ground.
6/5 = 51/2
=122.4 sec
h1 = 6 + 5 t
h2 = 51 - 2 t
when is h1 = h2 ?
6 + 5 t = 51 - 2 t
7 t = 45
t = 6 3/7 seconds
To solve this problem, we need to find the time at which both balloons are at the same height above the ground.
Let's assume that after t seconds, the first balloon will be at a height of h1 and the second balloon will be at a height of h2.
The height of the first balloon can be calculated using the equation h1 = 6 + 5t, because it started at a height of 6 feet and ascends at a rate of 5 feet per second.
The height of the second balloon can be calculated using the equation h2 = 51 - 2t, because it started at a height of 51 feet and descends at a rate of 2 feet per second.
Now, we want to find the time when h1 = h2. We can set up an equation:
6 + 5t = 51 - 2t
To solve this equation, we can first simplify it:
7t = 51 - 6
7t = 45
Now, divide both sides by 7 to isolate t:
t = 45/7
By dividing 45 by 7, we get:
t ≈ 6.43 seconds
Therefore, both balloons will be at the same height above the ground after approximately 6.43 seconds.