The motion of the stone can be modeled as a free fall; such that its velocity would be directly related to time, and its vertical displacement proportional to time squared- given that the free fall started from rest.
#"h"(t)=-1/2*"g"*t^2+v_0*t+"h"_0" "t \ge 0#
Where #"v"_0# the initial vertical velocity and #"h"_0# the initial height. The question implied that the stone is dropped with no vertical velocity (not given) such that #v_0=0#. Let the initial position be of height #"h"_0=0#. Hence
#"h"(t)=-1/2*"g"*t^2#
Assuming that #"g"=9.8 color(white)(l) "m"*"s"^(-2)#. Letting #"h"(t_f)=-122.5 color(white)(l)"m"#- the height of the water surface underneath with respect to the initial position of the stone- and solving for #t_f# gives
#t_f=sqrt(("h"(t_f))/(-1/2*"g"))=sqrt((-122.5)/(-1/2*9.8))=5.0color(grey)(0)color(white)(l)"s"#
That is: the collision takes place #5 color(white)(l)"seconds"# after the release of the stone. Note that it takes #t_"sound"=5.36-5.0color(grey)(0)=0.3 color(grey)(6)# more #"seconds"# for sound to cover the very distance #"s"_("sound")=122.5 color(white)(l)"m"#
before reaching the observer at #h_0#.
Letting the velocity of sound in air be #v_"sound"#,
#s_(sound)=v_"sound" *t_"sound"#
Hence #v_"sound"=(s_(sound))/(t_"sound")=(122.5 color(white)(l)m)/(0.3 color(grey)(6)color(white)(l)"s")=3.4*10^2 color(white)(l)"m"*"s"^(-1)#
