The rock's altitude y at time t, thrown with initial velocity v, is given by
[tex]y=74\,\mathrm m+vt-\dfrac12gt^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.
After t = 2.7 s, the rock reaches the water (0 altitude), so
[tex]0=74\,\mathrm m+v(2.7\,\mathrm s)-\dfrac12g(2.7\,\mathrm s)^2[/tex]
[tex]\implies v=-\dfrac{74\,\mathrm m-\frac g2(2.7\,\mathrm s)^2}{2.7\,\mathrm s}\approx-14.177\dfrac{\rm m}{\rm s}[/tex]
so the rock was thrown with a velocity with magnitude 14 m/s and downward direction.
Its velocity at time t is [tex]v-gt[/tex] (with no horizontal component), so that at the moment it hits the water, its velocity is
[tex]v-g(2.7\,\mathrm s)\approx-40.637\dfrac{\rm m}{\rm s}[/tex]
That is, its final velocity has an approximate magnitude of 41 m/s, also directed downward.
