a balloon is rising vertically at the rate of 2 m/s. an observer is standing on the ground 100m away from a point directly below the balloon. at what rate is the distance between the observer and the balloon changing when the balloon is 160m high?

the distance s of the balloon at height h is given by

s^2 = 100^2 + h^2

2s ds/dt = 2h dh/dt
we know dh/dt = 2, so when the balloon is 160m up,

2√(100^2+160^2) ds/dt = 2(160)(2)
ds/dt = 16/√89 m/s

To find the rate at which the distance between the observer and the balloon is changing, we can use the concept of related rates. This involves finding an equation that relates the variables involved and then differentiating it with respect to time.

Let's start by visualizing the scenario. We have a right triangle formed by the observer, the point directly below the balloon, and the balloon itself. The height of the balloon is the vertical leg of the triangle, and the distance between the observer and the point directly below the balloon is the horizontal leg. The hypotenuse represents the distance between the observer and the balloon.

Let's denote the height of the balloon as h, the distance between the observer and the balloon as x, and the distance between the observer and the point directly below the balloon as y.

From this, we can establish the equation of the right triangle:

x^2 + y^2 = h^2 ---(1)

Differentiating both sides of equation (1) with respect to time, we get:

2x(dx/dt) + 2y(dy/dt) = 2h(dh/dt) ---(2)

Now we need to find the values of x, y, and dh/dt when h = 160 m.

Given:
dh/dt = 2 m/s (the rate at which the height is changing)
h = 160 m
y = 100 m

To find x, we can use equation (1):

x^2 + y^2 = h^2

x^2 + (100m)^2 = (160m)^2
x^2 + 10000m^2 = 25600m^2
x^2 = 25600m^2 - 10000m^2
x^2 = 15600m^2
x = √(15600m^2)
x ≈ 124.99m

Now, substitute the values of x, y, h, and dh/dt into equation (2) and solve for dx/dt:

2x(dx/dt) + 2y(dy/dt) = 2h(dh/dt)

2(124.99m)(dx/dt) + 2(100m)(0) = 2(160m)(2m/s)
2(124.99m)(dx/dt) = 2(160m)(2m/s)
2(124.99m)(dx/dt) = 640m^2/s
(249.98m)(dx/dt) = 640m^2/s
dx/dt = 640m^2/s ÷ 249.98m
dx/dt ≈ 2.56m/s

Therefore, the rate at which the distance between the observer and the balloon is changing when the balloon is 160m high is approximately 2.56 m/s.