Answer
$70.5m/s$
Work Step by Step
We can find the required speed as follows:
$f^{\prime}=(\frac{1+u_{\circ}/v}{1+u_s/v})f$
This simplifies to:
$u_{\circ}=v[(\frac{f^{\prime}}{f})(1+\frac{u_s}{v})-1]$
We plug in the known values to obtain:
$u_{\circ}=(343)[(\frac{135}{124})(1+\frac{36.8}{343})-1]$
$u_{\circ}=70.5m/s$
