Question

Airplane at 19,200 feet descending at a rate of 40 feet per second. Another airplane takes off and ascends at a rate of 60 feet per second. After how many seconds will the airplanes be at the same height? What is the height

Answer 1

After 192 seconds, the two airplanes will be at the same height, which is 11,520 feet.

Step-by-step explanation:

To find the time it takes for the two airplanes to be at the same height, we need to set up an equation based on their rates of ascent and descent. Let's assume the height of the descending airplane is given by h1 and the height of the ascending airplane is given by h2. The descending airplane is descending at a rate of 40 feet per second, so its height after t seconds can be represented by the equation h1 = 19,200 - 40t. The ascending airplane is ascending at a rate of 60 feet per second, so its height after t seconds can be represented by the equation h2 = 60t. To find the time at which the two airplanes are at the same height, we can set h1 equal to h2 and solve for t: 19,200 - 40t = 60t. Simplifying this equation, we get 100t = 19,200, so t = 192 seconds. Therefore, after 192 seconds, the two airplanes will be at the same height.

To find the height at which the two airplanes are at, we can substitute the value of t into either the equation for h1 or h2. Let's use the equation for h1: h1 = 19,200 - 40(192) = 19,200 - 7,680 = 11,520 feet. Therefore, after 192 seconds, the two airplanes will be at a height of 11,520 feet.

Answer 2

Step-by-step explanation:

Let t = flying time in seconds for each plane

Descending rate = 40 ft/s

Ascending rate = 60 ft/s

Descending height, hd= 19200 - 40t

Ascending height, ha = 60t

Equating hd = ha, therefore:

60t = 19200-40t

60t + 40t = 19200

100t = 19200

t = 19200/100

t = 192 seconds

h = 19200 - 192(40)

h = 19200 - 7680

= 11,520 ft.