Assuming the cannonball is fired horizontally, its horizontal velocity stays at a constant 76 m/s. At the point it hits the ground, it has a speed of 89 m/s, so if its vertical velocity at that moment is [tex]v_y[/tex], we have
[tex]89\dfrac{\rm m}{\rm s}=\sqrt{\left(76\dfrac{\rm m}{\rm s}\right)^2+{v_y}^2}\implies{v_y}^2\approx2145\dfrac{\mathrm m^2}{\mathrm s^2}[/tex]
(taking the negative square root because we take the downward direction to be negative)
Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are the initial and final velocities, respectively; [tex]a[/tex] is the acceleration; and [tex]\Delta x[/tex] is the change in position of a body. In the cannonball's case, it starts with 0 vertical velocity and is subject to a downward acceleration with magnitude [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex]. So we have
[tex]2145\dfrac{\mathrm m^2}{\mathrm s^2}-0=-2g\Delta y\implies\Delta y\approx-109.44\,\mathrm m[/tex]
(which is negative because we take the cannonball's starting position at the top of the cliff to be the origin) so the cliff is about 109 m high.
