Question

An observer stands 200 m from the launch site of a hot-air balloon at an elevation equal to the elevation of the launch site. The balloon rises vertically at a constant rate of 4m/s. How fast is the angle of elevation of the balloon increasing 30 s after the launch? (The angle of elevation is the angle between the ground and the observer's line of sight to the balloon.)

please give me explanation of this question.

Answer 1

The angle of elevation of the balloon is increasing at a rate of approximately 0.02 radians per second 30 seconds after the launch.

Step-by-step explanation:

To solve this problem, we can use trigonometry and related rates. Let's start by defining the variables:

• Let x represent the horizontal distance between the observer and the balloon.
• Let y represent the height of the balloon above the observer.
• Let θ represent the angle of elevation of the balloon.

We are given that the balloon rises vertically at a constant rate of 4m/s. This means that the rate of change of y with respect to time (dy/dt) is 4 m/s.

We need to find how fast the angle of elevation of the balloon is changing with respect to time (dθ/dt).

By using trigonometric relationships, we can express tan(θ) as y/x. Taking the derivative of both sides with respect to time, we get:

sec^2(θ) * (dθ/dt) = (1/x) * (dy/dt)

Substituting the known values, we have:

sec^2(θ) * (dθ/dt) = (1/200) * 4

Simplifying, we find:

(dθ/dt) = (1/200) * 4 * sec^2(θ)

To find the value of (dθ/dt) 30 seconds after the launch, we need to find the value of θ at that time. Since the balloon is rising vertically, the height of the balloon above the observer is increasing at a constant rate of 4m/s. So, after 30 seconds, the height of the balloon above the observer would be y = 4 * 30 = 120 meters.

We can use the tangent function to find the angle of elevation:

tan(θ) = y/x

tan(θ) = 120/200

θ = tan-1(120/200)

θ ≈ 30.964 degrees

Now, substituting the value of θ, we can find (dθ/dt):

(dθ/dt) ≈ (1/200) * 4 * sec^2(30.964)

(dθ/dt) ≈ 0.02 radians per second

Answer 2

Step-by-step explanation:

This is a question of related rates

from the problem description , you should be able to see that

tan Φ = h / 200 where h is the height of the balloon

rearrange to h = 200 tan Φ

now differentiate with respect to 't'

dh / dt = 200 sec^2 Φ dΦ/dt

Now...when the balloon is 30 seconds into flight h = 120 m

then Φ at that point is arctan ( 120 /200) = 30.9637 degrees

dh/ dt is given as 4 m/s now just plug in and solve for dΦ/dt:

4 = 200 sec^2 (30.9637) dΦ/dt

dΦ/dt = .0147 radians/sec