## anonymous 5 years ago an observer stands 200 meters from the launch site of a hot-air balloon. balloon rises 4m/sec at a constant rate. how fast is the angle of elevation of the balloon increasing 30sec after the launch?

1. anonymous

after 30sec, the balloon reaches a height of 120m. the angle is then: $\alpha=\tan^{-1} (120/200) = 31degrees$

2. anonymous

1.36

3. anonymous

sorry about the 1.36- I had a bad connection and couldnt finish. $d \theta/dt=4m/(200\sec ^{2}\theta)$ and $\sec ^{2}\theta=1.36$