A hot air balloon rising vertically is tracked by an observer located 8 km from the lift-off point. At a certain moment, the angle between the observer's line of sight and the horizontal is (pi )/(5) and it is changing at a rate of 0.2rad/min How fast is the balloon rising at this moment? Let y be the height of the balloon (in km), t be time (in minutes), and Theta the angle between the line-of-sight and the horizontal (in radians). (Use decimal notation.Give your answer to two decimal places.) (d^2y)/(dt)approx square kwint

To determine how fast the balloon is rising at the given moment, we need to relate the height of the balloon to the angle and use the given rate of change of the angle.Given:- The observer is 8 km from the lift-off point.- The angle radians.- The rate of change of the angle rad/min.We can use trigonometry to relate , , and the distance from the observer to the lift-off point. Specifically, we use the tangent function: Differentiating both sides with respect to time : We need to find . First, let's calculate \( \sec^2(\Theta) \): Since \( \tan(\Theta) = \frac{y}{8} \), we have: Now, calculate \( \tan\left(\frac{\pi}{5}\right) \): Thus, Next, we calculate \( \sec^2\left(\frac{\pi}{5}\right) \): Now substitute these values into the differentiated equation: Solving for : Therefore, the balloon is rising at a rate of approximately km/min at that moment.