We know that it was launched with a speed #v#, and at its highest point, the #y#-velocity is zero, so the speed of #v/2# is solely the #x#-component (which does not change).
The initial #x#-velocity is thus also #v/2#, and the magnitude of the initial velocity is #v#, so the initial #y#-velocity is given by the Pythagorean theorem:
#v_(0y) = sqrt((v)^2 - (v/2)^2) = (vsqrt3)/2#
(which are components of a #30^"o"-60^"o"-90^"o"# triangle)
The maximum height #h# attained is given by the kinematics equation
#(v_y)^2 = (v_(0y))^2 - 2gh#
Plugging in known values, we have
#0 = 0.75v^2 - 2(9.81color(white)(l)"m/s"^2)h#
Rearranging and simplifying gives
#h = (0.75v^2)/(19.62color(white)(l)"m/s"^2)#
We can also simply write it with the acceleration #g# to make things even simpler:
#color(blue)(h = (0.75v^2)/(2g)#
or
#color(blue)(h = (3v^2)/(8g)#
