At the time the cannonball reaches the ground, the vertical height of the cannonball from the ground is 0, i.e. h = 0.
Thus, the time it will take the cannonball to reach the ground is given by
[tex]-16t^2+128t=0 \\ \\ \Rightarrow t(-16t+128)=0 \\ \\ \Rightarrow t=0\ or\ -16t+128=0 \\ \\ \Rightarrow16t=128 \\ \\ \Rightarrow t= \frac{128}{16} =8[/tex]
Therefore, it will take the cannonball 8 seconds to reach the ground.
