fluid dynamics calculating the maximum possible speed and finding the pressure At a point $A$ in the flow of a compressible fluid, the flow speed is $100$ms$^{-1}$ and the Mach number of the flow is $0.6$. Assuming the flow is adiabatic with index $\gamma = 1.4$, I need to find the following.
(i) the maximum possible flow speed on the streamline through $A$
(ii) the flow speed at a point where the pressure has fallen to half its value at the point $A$. 
I'm completely stumped on how to start this problem. I'm assuming I will need to use Bernoulli's equation but I am unsure. Could anyone please help me understand the methodology behind this type of problem? 
I know that $\left(\frac{\gamma}{\gamma-1}\frac{p}{\rho}\right)+\frac{v^{2}}{2}=$constant. From the comment, I got to 
$$ \frac{C_{s}^{2}}{\gamma-1}+\frac{C_{s}^{2}M^{2}}{2} = \textrm{constant} = C_{s}^{2}\left(\frac{1}{\gamma-1}+\frac{M^{2}}{2}\right) $$
but now how do I continue? Could I continue like  this? 
$$ C_{sA}^{2}\left(\frac{1}{\gamma-1}+\frac{M_{A}^{2}}{2}\right) = C_{sB}^{2}\left(\frac{1}{\gamma-1}+\frac{M_{B}^{2}}{2}\right) $$
Now I know $M=0.6$, $v=100$ms$^{-1}$, $\gamma=1.4$ but I don't know what to continue on from here? Here $C_{sA}$ is the speed of sound at $A$ and similarly for $C_{sB}$. 
 A: There are two extremes:
(1) Stagnation, where $v = v_0 = 0$, $p = p_0$, and $\rho= \rho_0$. This occurs, for example, at a stagnation points on an obstacle or in a large reservoir where the gas is quiescent and from which the gas escapes and expands.
(2) Maximum velocity, where $v = v_{\text{max}}$ and $p = 0$. This occurs when the temperature is zero and all thermal energy is converted to kinetic energy. This is what you are trying to calculate for part (i).
From Bernoulli's equation we have 
$$\tag{1}\frac{v^2}{2} + \frac{\gamma}{\gamma-1} \frac{p}{\rho} = \frac{\gamma}{\gamma-1} \frac{p_0}{\rho_0}.$$
Since $v = v_{\text{max}}$ when $p = 0$, it follows that 
$$\tag{2} v_{\text{max}} = \sqrt{\frac{2}{\gamma-1}}\sqrt{\frac{\gamma p_0}{\rho_0}} = \sqrt{\frac{2}{\gamma-1}}c_0.$$
We can solve for $c_0$, the speed of sound at stagnation conditions, as well as $v_{\text{max}}$, using the given data $v_A = 100$ and $M_A = 0.6$.  
Using $c_A^2 = \gamma p_A/ \rho_A$ and (1), we have
$$v_A^2 + \frac{2}{\gamma-1} c_A^2 = \frac{2}{\gamma-1} c_0^2\\ \implies v_A^2 + \frac{2}{\gamma-1} \frac{v_A^2}{M_A^2} = \frac{2}{\gamma-1} c_0^2 $$
whence, applying (2), we get the result
$$v_{\text{max}} = \sqrt{\frac{2}{\gamma-1}} c_0 = v_A \sqrt{1 + \frac{2}{(\gamma-1)M_A^2}}.$$
Completing part (ii) is straightforward from here using the adiabatic equation of state
$$\frac{p}{\rho^\gamma} = \text{constant}.$$
At point $B$ where $P_B = P_A/2$, we have 
$$\left(\frac{\rho_B}{\rho_A}\right)^\gamma = \frac{P_B}{P_A} = \frac{1}{2}\\ \implies \rho_B = \frac{\rho_A}{2^{1/\gamma}} \\ \tag{3}\implies \frac{P_B}{\rho_B} = \frac{1}{2^{1 - 1/\gamma}}\frac{P_A}{\rho_A}$$
Solve for $P_A/\rho_A$ using 
$$\frac{v_A^2}{2} + \frac{\gamma}{\gamma -1} \frac{P_A}{\rho_A} = \frac{v_{\text{max}}^2}{2},$$
and then solve for $v_B$ using (3) to find $P_B/\rho_B$ and 
$$\frac{v_B^2}{2} + \frac{\gamma}{\gamma -1} \frac{P_B}{\rho_B} = \frac{v_{\text{max}}^2}{2}.$$
