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

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(dx)/(dt)+2\text{y}(dy)/(\differentialDt t)=2\text{z}\frac{\text{dz}}{dt} \\ x(dx)/(dt)+\text{y}(dy)/(\differentialDt t)=z(dz)/(dt) \end{gathered}`
`\begin{gathered} (dx)/(\differentialDt t)=0 \\ (az)/(dt)=(y)/(z)(.dy)/(\differentialDt t) \\ =(50ft)/(158.118).(8ft)/(\sec ) \\ =(2.5ft)/(\sec ) \end{gathered}`

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