#### Answer

The speed of the cannon ball when it reaches the ground is 81.2 m/s

#### Work Step by Step

We can use conservation of energy to find the speed $v_f$ of the cannon ball when it reaches the ground. Therefore;
$KE_f+PE_f = KE_0+PE_0$
$\frac{1}{2}mv_f^2+0=\frac{1}{2}mv_0^2+mgh$
$v_f^2=v_0^2+2gh$
$v_f= \sqrt{v_0^2+2gh}$
$v_f= \sqrt{(80~m/s)^2+(2)(9.80~m/s^2)(10~m)}$
$v_f = 81.2~m/s$
The speed of the cannon ball when it reaches the ground is 81.2 m/s.