Question

A sounding body emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of 2 m/s, after one second when it started to fall. The difference in the frequency observed by the man in balloon just before and just after crossing the body will be(in Hz)
(Take velocity of sound as 300 m/s and g=10 $m/s^2$)

Answer 1

$f=f_0\left(\frac{\nu \pm {\nu }_0}{\nu \pm {\nu }_s}\right)Whenapproaching:f_a=150\left(\frac{300+2}{300-10}\right)Whenreceding:f_r=150\left(\frac{300-2}{300+10}\right)\Rightarrow f_a-f_r=12Hz$