an observer stands 200 m from the launch site of a hot-air balloon. the balloon rises vertically at a constant rate of 4 m/s. how fast is the angle of elevation increasing 30 s after the launch.
To find the rate at which the angle of elevation is increasing, we will need to use trigonometry. Let's denote the angle of elevation as θ.
Initially, when the balloon starts ascending, the observer is standing 200 m away from the launch site. This forms a right triangle where the hypotenuse is the line of sight from the observer to the balloon, the vertical side is the altitude of the balloon, and the horizontal side is the distance between the observer and the launch site.
We are given that the balloon rises vertically at a constant rate of 4 m/s. So, after 30 seconds, the balloon has risen by 4 m/s * 30 s = 120 m.
Now, we can calculate the distance between the observer and the balloon after 30 seconds by using the Pythagorean theorem. Let's call this distance x.
Using the Pythagorean theorem:
x^2 = 200^2 + (200 + 120)^2
x^2 = 40000 + 320^2
x^2 = 40000 + 102400
x^2 = 142400
x ≈ 377.49 m
Now, we can find the tangent of the angle of elevation by dividing the altitude (120 m) by the distance from the observer to the balloon (377.49 m).
tan(θ) = 120 / 377.49
θ ≈ 18.08 degrees
To find how fast the angle of elevation is changing, we differentiate θ with respect to time (t). Let's denote dθ/dt as the rate at which the angle of elevation is changing.
Differentiating both sides of the equation with respect to time:
d/dt(tan(θ)) = d/dt(120 / 377.49)
Using the chain rule of differentiation, we get:
sec^2(θ) * dθ/dt = (0 * 377.49 - 120 * (1/t^2)) / (377.49^2)
sec^2(θ) * dθ/dt = -120 / (377.49^2)
Plugging in the value of θ we found earlier:
sec^2(18.08) * dθ/dt = -120 / (377.49^2)
Using a calculator, we find:
sec^2(18.08) ≈ 1.0417
So, the equation becomes:
1.0417 * dθ/dt = -120 / (377.49^2)
Now, we can solve for dθ/dt:
dθ/dt ≈ -120 / (377.49^2 * 1.0417)
Evaluating this expression using a calculator, we find:
dθ/dt ≈ -0.00814 degrees per second
Therefore, the angle of elevation is decreasing at a rate of approximately 0.00814 degrees per second, 30 seconds after the launch.