If the distance becomes twice as much, the time for the stone to reach the water would $sqrt{2}$ times as much. However, the time required for the sound to go back to your each doubles. Add these to get the total time.

$t_{total}=sqrt{2}t_1+2t_2 ;<;2(t_1+t_2)$

Therefore, it is less than twice.

Using the following equation, calculate the final velocity

$v_f=sqrt{v_i^2+2gd}=sqrt{(2.10m/s)^2+2(9.81m/s^2)(7.35)}=12.2m/s$

Now, solve for $t_1$

$t_1=dfrac{v_f-v_i}{g}=dfrac{12.2;m/s-2.10;m/s}{9.81;m/s^2}=1.03;s$

Now, calculate for $t_2$

$t_2=dfrac{d}{v}=dfrac{7.35m}{343 m/s}=0.0214;s$

Get the total of $t_1$ and $t_2$

$$
t=t_1+t_2=1.03+0.0214=1.05s
$$

Calculate the time by dividing the distance by the velocity.

$t=dfrac{d}{v}=dfrac{95;m}{343;m/s}=0.277;s$

Now, compute for the speed of the cannonball using basic equations of motion

$$
v=v_i-gt=(44;m/s)-(9.81m/s^2)(0.277;s)=41m/s
$$