Question

A small balloon is released at a point 150 feet away from an observer who is on level ground. If the balloon goes straight up at the rate of 8 ft/sec, how fast is the distance from the observer to the balloon increasing when the balloon is 50 ft. high

Answer 1

Solution

How fast is the distance from the observer to the balloon:

Initial distance between observer and balloon is d = 150 ft;

the speed of the balloon is v = 8 ft/sec;

when balloon is h =50 ft high, final distance can be calculated as;

$\begin{gathered} d^2_f=h^2+d^2_0 \\ d_f=\sqrt[]{h^2+d^2_0} \\ d_f=\sqrt[]{150^2+50^2} \\ d_f=\sqrt[]{25000} \\ d_f=158.114ft \end{gathered}$

$\begin{gathered} \text{speed = }\frac{\text{distance}}{\text{time}} \\ time=\frac{dis\tan ce}{\text{speed}} \end{gathered}$$\begin{gathered} x^2+y^2=z^2 \\ 2x\frac{dx}{dt}+2\text{y}\frac{dy}{\differentialDt t}=2\text{z}\frac{\text{dz}}{dt} \\ x\frac{dx}{dt}+\text{y}\frac{dy}{\differentialDt t}=z\frac{dz}{dt} \end{gathered}$$\begin{gathered} \frac{dx}{\differentialDt t}=0 \\ \frac{az}{dt}=\frac{y}{z}\frac{.dy}{\differentialDt t} \\ =\frac{50ft}{158.118}.\frac{8ft}{\sec } \\ =\frac{2.5ft}{\sec } \end{gathered}$

Therefore the distance from the observer increasing = 2.5ft/sec