Answer
$v_{max}=2068ft/s$
Work Step by Step
The required maximum speed can be determined as follows:
$t=\frac{m_f}{q}=\frac{300}{15}=20s$
Now, $v_{max}=v_e\ln(\frac{W_r+W_f}{W_r+W_r-qt})$
We plug in the known values to obtain:
$v_{max}=4400\ln (\frac{500+300}{500+300-15(20)})$
This simplifies to:
$v_{max}=2068ft/s$
