If you launch the cannonball from the ground instead (hi=0), what does that change in the original equation?

Answer:
the value of hi changes from 58 to 0:
h(t) = -4.9t^2 +19.8t
Step-by-step explanation:
If the initial height changes, then the term in the equation that represents that height will change. The initial height term is hi, so it changes from 58 to 0.
[tex]h(t)=-4.9t^2+19.8t+0\\\\\boxed{h(t)=-4.9t^2+19.8t}[/tex]