Question

a hot air ballon rising straight up from a level field is tracked by a range finder 500ft from the lift-off point.At the moment the range finder's elevation angle is pi/4, the angle is increasing at the rate of .14rad/min.How fast is the balloon rising at the moment?

Answer 1

Answer on Question #41904, Physics, Mechanics | Kinematics | Dynamics

A hot air balloon rising straight up from a level field is tracked by a range finder 500ft from the lift-off point. At the moment the range finder's elevation angle is pi/4, the angle is increasing at the rate of .14rad/min. How fast is the balloon rising at the moment?

Solution

Let $h$ is the height of the balloon, $\theta$ is the angle at the range finder to the balloon. So

$$\tan \theta = \frac{h}{500}.$$

Now let's take the derivative thus:

$$\sec^2(\theta) \frac{d\theta}{dt} = \left(\frac{1}{500}\right) \frac{dh}{dt}$$

Solving for $\frac{dh}{dt}$ gives us:

$$\frac{dh}{dt} = 500 \sec^2(\theta) \frac{d\theta}{dt}.$$

When $\theta = \frac{\pi}{4}$ and $\frac{d\theta}{dt} = 0.14 \frac{rad}{min}$ the rate of change of the height of the balloon is

$$\frac{dh}{dt} = 500 ft \cdot \sec^2\left(\frac{\pi}{4}\right) \frac{d\theta}{dt} \cdot 0.14 \frac{rad}{min} = 500 \cdot (\sqrt{2})^2 \cdot 0.14 = 140 \frac{ft}{min}.$$

Answer: $140 \frac{ft}{min}$.

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